Is there a connection between $zeta(-1)$ and Ramanujan's calculation of the sum over $mathbb{N}$?
$begingroup$
Let me elaborate a little on the matter that I've been mulling over for a little while. This essentially concerns the summation of $1+2+3+...$, how it equals $-1/12$ (in a certain sense, obviously not normal circumstances), and how this seems to strangely coincide with the Riemann zeta function at $s=-1$.
I'm essentially concerned over whether this is a coincidence, or if there's something deeper at play here. It would certainly strike me as one heck of a coincidence.
I'll elaborate a bit for those unfamiliar with these matters...
The Value of $zeta(-1)$:
We know, for $s in mathbb{C}$, with $Re(s)>1$, we can define the Riemann zeta function by
$$zeta(s) = sum_{n=1}^infty frac{1}{n^s}$$
This can be analytically continued to the entirety of the complex plane, giving a definition of the function for $Re(s)<1, sneq1$:
$$zeta(s) = 2^s pi^{s-1} sinleft(frac{pi s}{2} right) Gamma (1-s) zeta(1-s)$$
Through this and the known result $zeta(2)=pi^2/6$ we can see that
$$zeta(-1) = 2^{-1} pi^{-1-1} sinleft(frac{pi (-1)}{2} right) Gamma (1-(-1)) zeta(1-(-1)) = frac{1}{2} left( frac{1}{pi^2} right)(-1)(1)left(frac{pi^2}{6}right)=frac{-1}{12}$$
We also know that people - often mistakenly - claim $1+2+3+4+...=-1/12$ through this result, since the summation of the natural numbers appears if you plug in $s=-1$ into the original sum, $sum n^{-s}$.
This obviously isn't true - mistake aside, at least in what we consider the "usual topology of $mathbb{R}$", the summation $1+2+3+...$ doesn't converge in
- The limit of its partial sums (which are the triangle numbers)
- The limit of the averages of its partial sums (Cesaro summation)
- Through Abel summation (per Wikipedia - I don't really know of this myself)
...and apparently a number of other summation methods (again, according to Wikipedia; I'm only familiar with the first two methods). Yet, interestingly...
Ramanujan's Summation of $1+2+3+...$:
Now, we know that in, the usual sense, we cannot assign a value to an obviously divergent series. For example, we cannot say $sum a_n = x$ if this summation is divergent, and then perform manipulations on that as if it had a value, and try to deduce whatever $x$ is.
This doesn't always have to hold, just in the sense of the usual topology of the real numbers. (At least, this is what I was told by Thomas Andrews in the comments of an semi-related question here - related in the sense that it dealt with an "obviously divergent" product. I honestly don't know much about topologies.) I take this to mean that we can assign values to these "obviously divergent in the usual sense" summations, but it changes the context, the framework in which we're working.
So Ramanujan was essentially doing that, in a certain sense - implicitly considering an alternate topology of the reals, in which the summation over the naturals could be assigned a value $c$, i.e.
$$c = sum_{n=1}^infty n = 1+2+3+4+...$$
and then manipulating $c$ to try and determine its value. (At least, that's what he might have been doing, I'm not sure. He could've just been ignoring the topologies altogether and just seeing what he could do. Either way...)
Subtracting $4c$ from $c$, he obtained
$$-3c = 1-2+3-4+5-6+...$$
This summation is the expansion of the power series for $(1+x)^{-2}$ and thus, taking $x=1$, Ramanujan obtains
$$-3c = frac{1}{(1+1)^2} = frac{1}{4} ;;; Rightarrow ;;; c = frac{-1}{12}$$
So, The Question, and What I've Found:
So, by using the zeta function, we can show that $zeta(-1) = -1/12$. Erroneously, people claim this as the sum of the natural numbers, but interestingly enough, Ramanujan showed that, if you can assign the divergent sum of the naturals a value, then it coincidentally is $-1/12$.
So, is there a connection between the Riemann zeta function - in particular its value at $s=-1$ - and Ramanujan's method of summing over the naturals? Is a mere coincidence? And in particular, why does Ramanujan's summation yield the same result?
I'm not really sure where to begin looking into this. Ramanujan's summation introduces the notion of a different topology on the reals, but I've never had a proper introduction into topology so I'm a bit loathe to go into learning a whole subject to understand a connection that might not even exist.
I did find that a Wikiepdia article on the summation $1+2+3+...$ kinda touches on a possible connection but I don't really buy it. (In the section on zeta function regularization, it suddenly starts considering the Riemann zeta function instead of the summation which seems more like a "similar but related" thing than establishing a proper connection.)
I mean, I understand where we're coming from in zeta function regularization, and I can sort of see it. We consider the summation over the naturals as a special case of the Riemann zeta function, then associate the value of the summation with the value of the zeta function there, or its analytic continuation if applicable. I can get that.
But it doesn't answer what I feel is the heart of my question:
I think more than anything, my big question is... Ramanujan's method relied nothing on "regularization" or "zeta functions" or any of this higher level stuff. He just assumed the summation had a value, and manipulated the summation to find that value. Yes, it broke some assumptions about the topology of the reals - that I myself don't understand well - yet somehow got the same answer that this regularization and all that would. How is it right? What is the connection?
The "obviously wrong" method somehow got the right answer. I mean, sure, it's right in the sense of the topology of the reals in which these manipulations are valid, but it's wrong in the usual sense. Yet it achieves the right answer.
So I think my biggest question is why it's correct. Coincidence? Perhaps the zeta function relies on likewise assumptions of topology that I've just not known about? I'm not really sure where to go from here; as it is my research is already broaching topics I know little to nothing about.
sequences-and-series riemann-zeta divergent-series natural-numbers zeta-functions
asked Dec 18 '18 at 5:33
Eevee TrainerEevee Trainer
5,7871936
$endgroup$
add a comment |
$begingroup$
Let me elaborate a little on the matter that I've been mulling over for a little while. This essentially concerns the summation of $1+2+3+...$, how it equals $-1/12$ (in a certain sense, obviously not normal circumstances), and how this seems to strangely coincide with the Riemann zeta function at $s=-1$.
I'm essentially concerned over whether this is a coincidence, or if there's something deeper at play here. It would certainly strike me as one heck of a coincidence.
I'll elaborate a bit for those unfamiliar with these matters...
The Value of $zeta(-1)$:
We know, for $s in mathbb{C}$, with $Re(s)>1$, we can define the Riemann zeta function by
$$zeta(s) = sum_{n=1}^infty frac{1}{n^s}$$
This can be analytically continued to the entirety of the complex plane, giving a definition of the function for $Re(s)<1, sneq1$:
$$zeta(s) = 2^s pi^{s-1} sinleft(frac{pi s}{2} right) Gamma (1-s) zeta(1-s)$$
Through this and the known result $zeta(2)=pi^2/6$ we can see that
$$zeta(-1) = 2^{-1} pi^{-1-1} sinleft(frac{pi (-1)}{2} right) Gamma (1-(-1)) zeta(1-(-1)) = frac{1}{2} left( frac{1}{pi^2} right)(-1)(1)left(frac{pi^2}{6}right)=frac{-1}{12}$$
We also know that people - often mistakenly - claim $1+2+3+4+...=-1/12$ through this result, since the summation of the natural numbers appears if you plug in $s=-1$ into the original sum, $sum n^{-s}$.
This obviously isn't true - mistake aside, at least in what we consider the "usual topology of $mathbb{R}$", the summation $1+2+3+...$ doesn't converge in
- The limit of its partial sums (which are the triangle numbers)
- The limit of the averages of its partial sums (Cesaro summation)
- Through Abel summation (per Wikipedia - I don't really know of this myself)
...and apparently a number of other summation methods (again, according to Wikipedia; I'm only familiar with the first two methods). Yet, interestingly...
Ramanujan's Summation of $1+2+3+...$:
Now, we know that in, the usual sense, we cannot assign a value to an obviously divergent series. For example, we cannot say $sum a_n = x$ if this summation is divergent, and then perform manipulations on that as if it had a value, and try to deduce whatever $x$ is.
This doesn't always have to hold, just in the sense of the usual topology of the real numbers. (At least, this is what I was told by Thomas Andrews in the comments of an semi-related question here - related in the sense that it dealt with an "obviously divergent" product. I honestly don't know much about topologies.) I take this to mean that we can assign values to these "obviously divergent in the usual sense" summations, but it changes the context, the framework in which we're working.
So Ramanujan was essentially doing that, in a certain sense - implicitly considering an alternate topology of the reals, in which the summation over the naturals could be assigned a value $c$, i.e.
$$c = sum_{n=1}^infty n = 1+2+3+4+...$$
and then manipulating $c$ to try and determine its value. (At least, that's what he might have been doing, I'm not sure. He could've just been ignoring the topologies altogether and just seeing what he could do. Either way...)
Subtracting $4c$ from $c$, he obtained
$$-3c = 1-2+3-4+5-6+...$$
This summation is the expansion of the power series for $(1+x)^{-2}$ and thus, taking $x=1$, Ramanujan obtains
$$-3c = frac{1}{(1+1)^2} = frac{1}{4} ;;; Rightarrow ;;; c = frac{-1}{12}$$
So, The Question, and What I've Found:
So, by using the zeta function, we can show that $zeta(-1) = -1/12$. Erroneously, people claim this as the sum of the natural numbers, but interestingly enough, Ramanujan showed that, if you can assign the divergent sum of the naturals a value, then it coincidentally is $-1/12$.
So, is there a connection between the Riemann zeta function - in particular its value at $s=-1$ - and Ramanujan's method of summing over the naturals? Is a mere coincidence? And in particular, why does Ramanujan's summation yield the same result?
I'm not really sure where to begin looking into this. Ramanujan's summation introduces the notion of a different topology on the reals, but I've never had a proper introduction into topology so I'm a bit loathe to go into learning a whole subject to understand a connection that might not even exist.
I did find that a Wikiepdia article on the summation $1+2+3+...$ kinda touches on a possible connection but I don't really buy it. (In the section on zeta function regularization, it suddenly starts considering the Riemann zeta function instead of the summation which seems more like a "similar but related" thing than establishing a proper connection.)
I mean, I understand where we're coming from in zeta function regularization, and I can sort of see it. We consider the summation over the naturals as a special case of the Riemann zeta function, then associate the value of the summation with the value of the zeta function there, or its analytic continuation if applicable. I can get that.
But it doesn't answer what I feel is the heart of my question:
I think more than anything, my big question is... Ramanujan's method relied nothing on "regularization" or "zeta functions" or any of this higher level stuff. He just assumed the summation had a value, and manipulated the summation to find that value. Yes, it broke some assumptions about the topology of the reals - that I myself don't understand well - yet somehow got the same answer that this regularization and all that would. How is it right? What is the connection?
The "obviously wrong" method somehow got the right answer. I mean, sure, it's right in the sense of the topology of the reals in which these manipulations are valid, but it's wrong in the usual sense. Yet it achieves the right answer.
So I think my biggest question is why it's correct. Coincidence? Perhaps the zeta function relies on likewise assumptions of topology that I've just not known about? I'm not really sure where to go from here; as it is my research is already broaching topics I know little to nothing about.
sequences-and-series riemann-zeta divergent-series natural-numbers zeta-functions
asked Dec 18 '18 at 5:33
Eevee TrainerEevee Trainer
5,7871936
$endgroup$
$begingroup$
Nitpick: $zeta(s) = 2^s pi^{s-1} sinleft(frac{pi s}{2} right) Gamma (1-s) zeta(1-s)$ only gives $zeta(s)$ for $operatorname{Re}s<0$ if you start with the standard series. For the critical strip you need more.
$endgroup$
– Arthur
Dec 18 '18 at 7:33
$begingroup$
This "equality" occurs over and over again. At least in this question, it is highlighted that it does not hold in the usual sense
$endgroup$
– Peter
Dec 22 '18 at 14:38
add a comment |
$begingroup$
Let me elaborate a little on the matter that I've been mulling over for a little while. This essentially concerns the summation of $1+2+3+...$, how it equals $-1/12$ (in a certain sense, obviously not normal circumstances), and how this seems to strangely coincide with the Riemann zeta function at $s=-1$.
I'm essentially concerned over whether this is a coincidence, or if there's something deeper at play here. It would certainly strike me as one heck of a coincidence.
I'll elaborate a bit for those unfamiliar with these matters...
The Value of $zeta(-1)$:
We know, for $s in mathbb{C}$, with $Re(s)>1$, we can define the Riemann zeta function by
$$zeta(s) = sum_{n=1}^infty frac{1}{n^s}$$
This can be analytically continued to the entirety of the complex plane, giving a definition of the function for $Re(s)<1, sneq1$:
$$zeta(s) = 2^s pi^{s-1} sinleft(frac{pi s}{2} right) Gamma (1-s) zeta(1-s)$$
Through this and the known result $zeta(2)=pi^2/6$ we can see that
$$zeta(-1) = 2^{-1} pi^{-1-1} sinleft(frac{pi (-1)}{2} right) Gamma (1-(-1)) zeta(1-(-1)) = frac{1}{2} left( frac{1}{pi^2} right)(-1)(1)left(frac{pi^2}{6}right)=frac{-1}{12}$$
We also know that people - often mistakenly - claim $1+2+3+4+...=-1/12$ through this result, since the summation of the natural numbers appears if you plug in $s=-1$ into the original sum, $sum n^{-s}$.
This obviously isn't true - mistake aside, at least in what we consider the "usual topology of $mathbb{R}$", the summation $1+2+3+...$ doesn't converge in
- The limit of its partial sums (which are the triangle numbers)
- The limit of the averages of its partial sums (Cesaro summation)
- Through Abel summation (per Wikipedia - I don't really know of this myself)
...and apparently a number of other summation methods (again, according to Wikipedia; I'm only familiar with the first two methods). Yet, interestingly...
Ramanujan's Summation of $1+2+3+...$:
Now, we know that in, the usual sense, we cannot assign a value to an obviously divergent series. For example, we cannot say $sum a_n = x$ if this summation is divergent, and then perform manipulations on that as if it had a value, and try to deduce whatever $x$ is.
This doesn't always have to hold, just in the sense of the usual topology of the real numbers. (At least, this is what I was told by Thomas Andrews in the comments of an semi-related question here - related in the sense that it dealt with an "obviously divergent" product. I honestly don't know much about topologies.) I take this to mean that we can assign values to these "obviously divergent in the usual sense" summations, but it changes the context, the framework in which we're working.
So Ramanujan was essentially doing that, in a certain sense - implicitly considering an alternate topology of the reals, in which the summation over the naturals could be assigned a value $c$, i.e.
$$c = sum_{n=1}^infty n = 1+2+3+4+...$$
and then manipulating $c$ to try and determine its value. (At least, that's what he might have been doing, I'm not sure. He could've just been ignoring the topologies altogether and just seeing what he could do. Either way...)
Subtracting $4c$ from $c$, he obtained
$$-3c = 1-2+3-4+5-6+...$$
This summation is the expansion of the power series for $(1+x)^{-2}$ and thus, taking $x=1$, Ramanujan obtains
$$-3c = frac{1}{(1+1)^2} = frac{1}{4} ;;; Rightarrow ;;; c = frac{-1}{12}$$
So, The Question, and What I've Found:
So, by using the zeta function, we can show that $zeta(-1) = -1/12$. Erroneously, people claim this as the sum of the natural numbers, but interestingly enough, Ramanujan showed that, if you can assign the divergent sum of the naturals a value, then it coincidentally is $-1/12$.
So, is there a connection between the Riemann zeta function - in particular its value at $s=-1$ - and Ramanujan's method of summing over the naturals? Is a mere coincidence? And in particular, why does Ramanujan's summation yield the same result?
I'm not really sure where to begin looking into this. Ramanujan's summation introduces the notion of a different topology on the reals, but I've never had a proper introduction into topology so I'm a bit loathe to go into learning a whole subject to understand a connection that might not even exist.
I did find that a Wikiepdia article on the summation $1+2+3+...$ kinda touches on a possible connection but I don't really buy it. (In the section on zeta function regularization, it suddenly starts considering the Riemann zeta function instead of the summation which seems more like a "similar but related" thing than establishing a proper connection.)
I mean, I understand where we're coming from in zeta function regularization, and I can sort of see it. We consider the summation over the naturals as a special case of the Riemann zeta function, then associate the value of the summation with the value of the zeta function there, or its analytic continuation if applicable. I can get that.
But it doesn't answer what I feel is the heart of my question:
I think more than anything, my big question is... Ramanujan's method relied nothing on "regularization" or "zeta functions" or any of this higher level stuff. He just assumed the summation had a value, and manipulated the summation to find that value. Yes, it broke some assumptions about the topology of the reals - that I myself don't understand well - yet somehow got the same answer that this regularization and all that would. How is it right? What is the connection?
The "obviously wrong" method somehow got the right answer. I mean, sure, it's right in the sense of the topology of the reals in which these manipulations are valid, but it's wrong in the usual sense. Yet it achieves the right answer.
So I think my biggest question is why it's correct. Coincidence? Perhaps the zeta function relies on likewise assumptions of topology that I've just not known about? I'm not really sure where to go from here; as it is my research is already broaching topics I know little to nothing about.
sequences-and-series riemann-zeta divergent-series natural-numbers zeta-functions
asked Dec 18 '18 at 5:33
Eevee TrainerEevee Trainer
5,7871936
$endgroup$
Let me elaborate a little on the matter that I've been mulling over for a little while. This essentially concerns the summation of $1+2+3+...$, how it equals $-1/12$ (in a certain sense, obviously not normal circumstances), and how this seems to strangely coincide with the Riemann zeta function at $s=-1$.
I'm essentially concerned over whether this is a coincidence, or if there's something deeper at play here. It would certainly strike me as one heck of a coincidence.
I'll elaborate a bit for those unfamiliar with these matters...
The Value of $zeta(-1)$:
We know, for $s in mathbb{C}$, with $Re(s)>1$, we can define the Riemann zeta function by
$$zeta(s) = sum_{n=1}^infty frac{1}{n^s}$$
This can be analytically continued to the entirety of the complex plane, giving a definition of the function for $Re(s)<1, sneq1$:
$$zeta(s) = 2^s pi^{s-1} sinleft(frac{pi s}{2} right) Gamma (1-s) zeta(1-s)$$
Through this and the known result $zeta(2)=pi^2/6$ we can see that
$$zeta(-1) = 2^{-1} pi^{-1-1} sinleft(frac{pi (-1)}{2} right) Gamma (1-(-1)) zeta(1-(-1)) = frac{1}{2} left( frac{1}{pi^2} right)(-1)(1)left(frac{pi^2}{6}right)=frac{-1}{12}$$
We also know that people - often mistakenly - claim $1+2+3+4+...=-1/12$ through this result, since the summation of the natural numbers appears if you plug in $s=-1$ into the original sum, $sum n^{-s}$.
This obviously isn't true - mistake aside, at least in what we consider the "usual topology of $mathbb{R}$", the summation $1+2+3+...$ doesn't converge in
- The limit of its partial sums (which are the triangle numbers)
- The limit of the averages of its partial sums (Cesaro summation)
- Through Abel summation (per Wikipedia - I don't really know of this myself)
...and apparently a number of other summation methods (again, according to Wikipedia; I'm only familiar with the first two methods). Yet, interestingly...
Ramanujan's Summation of $1+2+3+...$:
Now, we know that in, the usual sense, we cannot assign a value to an obviously divergent series. For example, we cannot say $sum a_n = x$ if this summation is divergent, and then perform manipulations on that as if it had a value, and try to deduce whatever $x$ is.
This doesn't always have to hold, just in the sense of the usual topology of the real numbers. (At least, this is what I was told by Thomas Andrews in the comments of an semi-related question here - related in the sense that it dealt with an "obviously divergent" product. I honestly don't know much about topologies.) I take this to mean that we can assign values to these "obviously divergent in the usual sense" summations, but it changes the context, the framework in which we're working.
So Ramanujan was essentially doing that, in a certain sense - implicitly considering an alternate topology of the reals, in which the summation over the naturals could be assigned a value $c$, i.e.
$$c = sum_{n=1}^infty n = 1+2+3+4+...$$
and then manipulating $c$ to try and determine its value. (At least, that's what he might have been doing, I'm not sure. He could've just been ignoring the topologies altogether and just seeing what he could do. Either way...)
Subtracting $4c$ from $c$, he obtained
$$-3c = 1-2+3-4+5-6+...$$
This summation is the expansion of the power series for $(1+x)^{-2}$ and thus, taking $x=1$, Ramanujan obtains
$$-3c = frac{1}{(1+1)^2} = frac{1}{4} ;;; Rightarrow ;;; c = frac{-1}{12}$$
So, The Question, and What I've Found:
So, by using the zeta function, we can show that $zeta(-1) = -1/12$. Erroneously, people claim this as the sum of the natural numbers, but interestingly enough, Ramanujan showed that, if you can assign the divergent sum of the naturals a value, then it coincidentally is $-1/12$.
So, is there a connection between the Riemann zeta function - in particular its value at $s=-1$ - and Ramanujan's method of summing over the naturals? Is a mere coincidence? And in particular, why does Ramanujan's summation yield the same result?
I'm not really sure where to begin looking into this. Ramanujan's summation introduces the notion of a different topology on the reals, but I've never had a proper introduction into topology so I'm a bit loathe to go into learning a whole subject to understand a connection that might not even exist.
I did find that a Wikiepdia article on the summation $1+2+3+...$ kinda touches on a possible connection but I don't really buy it. (In the section on zeta function regularization, it suddenly starts considering the Riemann zeta function instead of the summation which seems more like a "similar but related" thing than establishing a proper connection.)
I mean, I understand where we're coming from in zeta function regularization, and I can sort of see it. We consider the summation over the naturals as a special case of the Riemann zeta function, then associate the value of the summation with the value of the zeta function there, or its analytic continuation if applicable. I can get that.
But it doesn't answer what I feel is the heart of my question:
I think more than anything, my big question is... Ramanujan's method relied nothing on "regularization" or "zeta functions" or any of this higher level stuff. He just assumed the summation had a value, and manipulated the summation to find that value. Yes, it broke some assumptions about the topology of the reals - that I myself don't understand well - yet somehow got the same answer that this regularization and all that would. How is it right? What is the connection?
The "obviously wrong" method somehow got the right answer. I mean, sure, it's right in the sense of the topology of the reals in which these manipulations are valid, but it's wrong in the usual sense. Yet it achieves the right answer.
So I think my biggest question is why it's correct. Coincidence? Perhaps the zeta function relies on likewise assumptions of topology that I've just not known about? I'm not really sure where to go from here; as it is my research is already broaching topics I know little to nothing about.
sequences-and-series riemann-zeta divergent-series natural-numbers zeta-functions
sequences-and-series riemann-zeta divergent-series natural-numbers zeta-functions
asked Dec 18 '18 at 5:33
Eevee TrainerEevee Trainer
5,7871936
asked Dec 18 '18 at 5:33
Eevee TrainerEevee Trainer
5,7871936
asked Dec 18 '18 at 5:33
Eevee TrainerEevee Trainer
5,7871936
asked Dec 18 '18 at 5:33
Eevee TrainerEevee Trainer
5,7871936
asked Dec 18 '18 at 5:33
Eevee TrainerEevee Trainer
5,7871936
5,7871936
$begingroup$
Nitpick: $zeta(s) = 2^s pi^{s-1} sinleft(frac{pi s}{2} right) Gamma (1-s) zeta(1-s)$ only gives $zeta(s)$ for $operatorname{Re}s<0$ if you start with the standard series. For the critical strip you need more.
$endgroup$
– Arthur
Dec 18 '18 at 7:33
$begingroup$
This "equality" occurs over and over again. At least in this question, it is highlighted that it does not hold in the usual sense
$endgroup$
– Peter
Dec 22 '18 at 14:38
add a comment |
$begingroup$
Nitpick: $zeta(s) = 2^s pi^{s-1} sinleft(frac{pi s}{2} right) Gamma (1-s) zeta(1-s)$ only gives $zeta(s)$ for $operatorname{Re}s<0$ if you start with the standard series. For the critical strip you need more.
$endgroup$
– Arthur
Dec 18 '18 at 7:33
$begingroup$
This "equality" occurs over and over again. At least in this question, it is highlighted that it does not hold in the usual sense
$endgroup$
– Peter
Dec 22 '18 at 14:38
$begingroup$
Nitpick: $zeta(s) = 2^s pi^{s-1} sinleft(frac{pi s}{2} right) Gamma (1-s) zeta(1-s)$ only gives $zeta(s)$ for $operatorname{Re}s<0$ if you start with the standard series. For the critical strip you need more.
$endgroup$
– Arthur
Dec 18 '18 at 7:33
$begingroup$
Nitpick: $zeta(s) = 2^s pi^{s-1} sinleft(frac{pi s}{2} right) Gamma (1-s) zeta(1-s)$ only gives $zeta(s)$ for $operatorname{Re}s<0$ if you start with the standard series. For the critical strip you need more.
$endgroup$
– Arthur
Dec 18 '18 at 7:33
$begingroup$
This "equality" occurs over and over again. At least in this question, it is highlighted that it does not hold in the usual sense
$endgroup$
– Peter
Dec 22 '18 at 14:38
$begingroup$
This "equality" occurs over and over again. At least in this question, it is highlighted that it does not hold in the usual sense
$endgroup$
– Peter
Dec 22 '18 at 14:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Well, "Ramanujan summation" usually refers to something else entirely, so let's not use that phrase here. Alternate topologies on the real numbers are another red herring; they're not relevant to this discussion. We're not evaluating the limit of a sequence of partial sums in any topology. With that stuff out of the way...
There are multiple ways to compute $zeta(-1)$. You've outlined using the functional equation to relate it to $zeta(2)$. But the more relevant method is in the Wikipedia article:
$$-3zeta(-1)=eta(-1)=lim_{xto 1^-}left(1-2x+3x^2-4x^3+cdotsright)=lim_{xto 1^-}frac{1}{(1+x)^2}=frac14$$
where $eta$ is the Dirichlet eta function, not to be confused with the unrelated and better-known Dedekind eta function.
This is a direct analogue to what Ramanujan wrote in his notebook:
$$-3c=1-2+3-4+mathrm{&c}=frac{1}{(1+1)^2}=frac14$$
Imagine, for the sake of argument, that Ramanujan's intention is in fact to compute $zeta(-1)$. Then every single step in his computation of "$c$" corresponds to a step in the computation of $zeta(-1)$, just with the variables $s$ and $x$ elided. And the computation of $zeta(-1)$ is rigorous; each step is justified in the Wikipedia article, which I'll quote for completeness:
Where both Dirichlet series converge, one has the identities:
$$
begin{alignat}{7}
zeta(s)&{}={}&1^{-s}+2^{-s}&&{}+3^{-s}+4^{-s}&&{}+5^{-s}+6^{-s}+cdots& \
2times2^{-s}zeta(s)&{}={}& 2times2^{-s}&& {}+2times4^{-s}&&{} +2times6^{-s}+cdots& \
left(1-2^{1-s}right)zeta(s)&{}={}&1^{-s}-2^{-s}&&{}+3^{-s}-4^{-s}&&{}+5^{-s}-6^{-s}+cdots&=eta(s) \
end{alignat}
$$
The identity $(1-2^{1-s})zeta(s)=eta(s)$ continues to hold when both functions are extended by analytic continuation to include values of $s$ for which the above series diverge. Substituting $s = −1$, one gets $−3zeta(−1) = eta(−1)$. Now, computing $eta(−1)$ is an easier task, as the eta function is equal to the Abel sum of its defining series, which is a one-sided limit.
And that explains why Ramanujan's manipulation yields the same value as $zeta(-1)$.
Perhaps the most obscure step in the argument is the usage of Abel summation to evaluate a Dirichlet series. The reference to Knopp justifies that step better than I would be able to. Beyond that, I'm not sure what else you're likely to object to. Feel free to ask about any particular step!
answered Dec 18 '18 at 7:35
Chris CulterChris Culter
20k43483
$endgroup$
1
$begingroup$
That "the eta function is equal to the Abel sum of its defining series" (so you can replace $lim_{s to -k}eta(s)$ by $lim_{z to 1} sum_n (-1)^{n+1} n^k z^n$) is not obvious : it is a consequence of $sum_{n=1}^N (-1)^{n+1} = frac{1+(-1)^{N+1}}{2}$ that summing by parts $K > -Re(s)$ times makes $sum_n (-1)^{n+1} n^{-s} z^n$ continuous in $|z| le 1, z ne -1$ and analytic in $s, Re(s) > -K$, whence $eta(s) =lim_{z to 1} sum_n (-1)^{n+1} n^{-s} z^n$ is valid for every $s$.
$endgroup$
– reuns
Dec 18 '18 at 7:43
$begingroup$
@reuns I do want to recognize that the equality isn't obvious! But it isn't an accident, either. The last time I read the Knopp reference, I got the impression that this phenomenon is not peculiar to this series. Rather, as a general principle, Abel summation works as well as you could possibly hope that it might for Dirichlet series. And there is some rigorous formulation of that last sentence. Am I off base there?
$endgroup$
– Chris Culter
Dec 18 '18 at 7:53
$begingroup$
It doesn't work with $sum_n n^{-s}$, only with Dirichlet series that summing by parts yields an analytic continuation
$endgroup$
– reuns
Dec 18 '18 at 8:02
$begingroup$
@reuns Sure, that series is too much to hope for. I meant "work" in the more lax sense of, where the Abel sum does exist, it agrees with the analytic continuation. That's all we need.
$endgroup$
– Chris Culter
Dec 18 '18 at 8:07
1
$begingroup$
Let $f(t) = sum_{n=1}^infty a_n e^{-nt}$ with $a_n = O(n^k)$ and $F(s) = sum_{n=1}^infty a_n n^{-s}, Re(s) > k+1$. Then $int_0^infty t^{s-1}f(t)dt = Gamma(s) F(s)$ and if $lim_{t to 0^+} f(t) = C ne 0$ converges then $int_0^infty t^{s-1}f(t)dt$ converges and is analytic for $Re(s) > 0$, obtaining that (the continuation of) $F(s)$ is analytic and $lim_{s to 0^+} F(s)=C$. But summing by parts $sum_{n=1}^infty a_n n^{-s}$ would have given the same result.
$endgroup$
– reuns
Dec 18 '18 at 8:40
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044810%2fis-there-a-connection-between-zeta-1-and-ramanujans-calculation-of-the-sum%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well, "Ramanujan summation" usually refers to something else entirely, so let's not use that phrase here. Alternate topologies on the real numbers are another red herring; they're not relevant to this discussion. We're not evaluating the limit of a sequence of partial sums in any topology. With that stuff out of the way...
There are multiple ways to compute $zeta(-1)$. You've outlined using the functional equation to relate it to $zeta(2)$. But the more relevant method is in the Wikipedia article:
$$-3zeta(-1)=eta(-1)=lim_{xto 1^-}left(1-2x+3x^2-4x^3+cdotsright)=lim_{xto 1^-}frac{1}{(1+x)^2}=frac14$$
where $eta$ is the Dirichlet eta function, not to be confused with the unrelated and better-known Dedekind eta function.
This is a direct analogue to what Ramanujan wrote in his notebook:
$$-3c=1-2+3-4+mathrm{&c}=frac{1}{(1+1)^2}=frac14$$
Imagine, for the sake of argument, that Ramanujan's intention is in fact to compute $zeta(-1)$. Then every single step in his computation of "$c$" corresponds to a step in the computation of $zeta(-1)$, just with the variables $s$ and $x$ elided. And the computation of $zeta(-1)$ is rigorous; each step is justified in the Wikipedia article, which I'll quote for completeness:
Where both Dirichlet series converge, one has the identities:
$$
begin{alignat}{7}
zeta(s)&{}={}&1^{-s}+2^{-s}&&{}+3^{-s}+4^{-s}&&{}+5^{-s}+6^{-s}+cdots& \
2times2^{-s}zeta(s)&{}={}& 2times2^{-s}&& {}+2times4^{-s}&&{} +2times6^{-s}+cdots& \
left(1-2^{1-s}right)zeta(s)&{}={}&1^{-s}-2^{-s}&&{}+3^{-s}-4^{-s}&&{}+5^{-s}-6^{-s}+cdots&=eta(s) \
end{alignat}
$$
The identity $(1-2^{1-s})zeta(s)=eta(s)$ continues to hold when both functions are extended by analytic continuation to include values of $s$ for which the above series diverge. Substituting $s = −1$, one gets $−3zeta(−1) = eta(−1)$. Now, computing $eta(−1)$ is an easier task, as the eta function is equal to the Abel sum of its defining series, which is a one-sided limit.
And that explains why Ramanujan's manipulation yields the same value as $zeta(-1)$.
Perhaps the most obscure step in the argument is the usage of Abel summation to evaluate a Dirichlet series. The reference to Knopp justifies that step better than I would be able to. Beyond that, I'm not sure what else you're likely to object to. Feel free to ask about any particular step!
answered Dec 18 '18 at 7:35
Chris CulterChris Culter
20k43483
$endgroup$
1
$begingroup$
That "the eta function is equal to the Abel sum of its defining series" (so you can replace $lim_{s to -k}eta(s)$ by $lim_{z to 1} sum_n (-1)^{n+1} n^k z^n$) is not obvious : it is a consequence of $sum_{n=1}^N (-1)^{n+1} = frac{1+(-1)^{N+1}}{2}$ that summing by parts $K > -Re(s)$ times makes $sum_n (-1)^{n+1} n^{-s} z^n$ continuous in $|z| le 1, z ne -1$ and analytic in $s, Re(s) > -K$, whence $eta(s) =lim_{z to 1} sum_n (-1)^{n+1} n^{-s} z^n$ is valid for every $s$.
$endgroup$
– reuns
Dec 18 '18 at 7:43
$begingroup$
@reuns I do want to recognize that the equality isn't obvious! But it isn't an accident, either. The last time I read the Knopp reference, I got the impression that this phenomenon is not peculiar to this series. Rather, as a general principle, Abel summation works as well as you could possibly hope that it might for Dirichlet series. And there is some rigorous formulation of that last sentence. Am I off base there?
$endgroup$
– Chris Culter
Dec 18 '18 at 7:53
$begingroup$
It doesn't work with $sum_n n^{-s}$, only with Dirichlet series that summing by parts yields an analytic continuation
$endgroup$
– reuns
Dec 18 '18 at 8:02
$begingroup$
@reuns Sure, that series is too much to hope for. I meant "work" in the more lax sense of, where the Abel sum does exist, it agrees with the analytic continuation. That's all we need.
$endgroup$
– Chris Culter
Dec 18 '18 at 8:07
1
$begingroup$
Let $f(t) = sum_{n=1}^infty a_n e^{-nt}$ with $a_n = O(n^k)$ and $F(s) = sum_{n=1}^infty a_n n^{-s}, Re(s) > k+1$. Then $int_0^infty t^{s-1}f(t)dt = Gamma(s) F(s)$ and if $lim_{t to 0^+} f(t) = C ne 0$ converges then $int_0^infty t^{s-1}f(t)dt$ converges and is analytic for $Re(s) > 0$, obtaining that (the continuation of) $F(s)$ is analytic and $lim_{s to 0^+} F(s)=C$. But summing by parts $sum_{n=1}^infty a_n n^{-s}$ would have given the same result.
$endgroup$
– reuns
Dec 18 '18 at 8:40
add a comment |
$begingroup$
Well, "Ramanujan summation" usually refers to something else entirely, so let's not use that phrase here. Alternate topologies on the real numbers are another red herring; they're not relevant to this discussion. We're not evaluating the limit of a sequence of partial sums in any topology. With that stuff out of the way...
There are multiple ways to compute $zeta(-1)$. You've outlined using the functional equation to relate it to $zeta(2)$. But the more relevant method is in the Wikipedia article:
$$-3zeta(-1)=eta(-1)=lim_{xto 1^-}left(1-2x+3x^2-4x^3+cdotsright)=lim_{xto 1^-}frac{1}{(1+x)^2}=frac14$$
where $eta$ is the Dirichlet eta function, not to be confused with the unrelated and better-known Dedekind eta function.
This is a direct analogue to what Ramanujan wrote in his notebook:
$$-3c=1-2+3-4+mathrm{&c}=frac{1}{(1+1)^2}=frac14$$
Imagine, for the sake of argument, that Ramanujan's intention is in fact to compute $zeta(-1)$. Then every single step in his computation of "$c$" corresponds to a step in the computation of $zeta(-1)$, just with the variables $s$ and $x$ elided. And the computation of $zeta(-1)$ is rigorous; each step is justified in the Wikipedia article, which I'll quote for completeness:
Where both Dirichlet series converge, one has the identities:
$$
begin{alignat}{7}
zeta(s)&{}={}&1^{-s}+2^{-s}&&{}+3^{-s}+4^{-s}&&{}+5^{-s}+6^{-s}+cdots& \
2times2^{-s}zeta(s)&{}={}& 2times2^{-s}&& {}+2times4^{-s}&&{} +2times6^{-s}+cdots& \
left(1-2^{1-s}right)zeta(s)&{}={}&1^{-s}-2^{-s}&&{}+3^{-s}-4^{-s}&&{}+5^{-s}-6^{-s}+cdots&=eta(s) \
end{alignat}
$$
The identity $(1-2^{1-s})zeta(s)=eta(s)$ continues to hold when both functions are extended by analytic continuation to include values of $s$ for which the above series diverge. Substituting $s = −1$, one gets $−3zeta(−1) = eta(−1)$. Now, computing $eta(−1)$ is an easier task, as the eta function is equal to the Abel sum of its defining series, which is a one-sided limit.
And that explains why Ramanujan's manipulation yields the same value as $zeta(-1)$.
Perhaps the most obscure step in the argument is the usage of Abel summation to evaluate a Dirichlet series. The reference to Knopp justifies that step better than I would be able to. Beyond that, I'm not sure what else you're likely to object to. Feel free to ask about any particular step!
answered Dec 18 '18 at 7:35
Chris CulterChris Culter
20k43483
$endgroup$
1
$begingroup$
That "the eta function is equal to the Abel sum of its defining series" (so you can replace $lim_{s to -k}eta(s)$ by $lim_{z to 1} sum_n (-1)^{n+1} n^k z^n$) is not obvious : it is a consequence of $sum_{n=1}^N (-1)^{n+1} = frac{1+(-1)^{N+1}}{2}$ that summing by parts $K > -Re(s)$ times makes $sum_n (-1)^{n+1} n^{-s} z^n$ continuous in $|z| le 1, z ne -1$ and analytic in $s, Re(s) > -K$, whence $eta(s) =lim_{z to 1} sum_n (-1)^{n+1} n^{-s} z^n$ is valid for every $s$.
$endgroup$
– reuns
Dec 18 '18 at 7:43
$begingroup$
@reuns I do want to recognize that the equality isn't obvious! But it isn't an accident, either. The last time I read the Knopp reference, I got the impression that this phenomenon is not peculiar to this series. Rather, as a general principle, Abel summation works as well as you could possibly hope that it might for Dirichlet series. And there is some rigorous formulation of that last sentence. Am I off base there?
$endgroup$
– Chris Culter
Dec 18 '18 at 7:53
$begingroup$
It doesn't work with $sum_n n^{-s}$, only with Dirichlet series that summing by parts yields an analytic continuation
$endgroup$
– reuns
Dec 18 '18 at 8:02
$begingroup$
@reuns Sure, that series is too much to hope for. I meant "work" in the more lax sense of, where the Abel sum does exist, it agrees with the analytic continuation. That's all we need.
$endgroup$
– Chris Culter
Dec 18 '18 at 8:07
1
$begingroup$
Let $f(t) = sum_{n=1}^infty a_n e^{-nt}$ with $a_n = O(n^k)$ and $F(s) = sum_{n=1}^infty a_n n^{-s}, Re(s) > k+1$. Then $int_0^infty t^{s-1}f(t)dt = Gamma(s) F(s)$ and if $lim_{t to 0^+} f(t) = C ne 0$ converges then $int_0^infty t^{s-1}f(t)dt$ converges and is analytic for $Re(s) > 0$, obtaining that (the continuation of) $F(s)$ is analytic and $lim_{s to 0^+} F(s)=C$. But summing by parts $sum_{n=1}^infty a_n n^{-s}$ would have given the same result.
$endgroup$
– reuns
Dec 18 '18 at 8:40
add a comment |
$begingroup$
Well, "Ramanujan summation" usually refers to something else entirely, so let's not use that phrase here. Alternate topologies on the real numbers are another red herring; they're not relevant to this discussion. We're not evaluating the limit of a sequence of partial sums in any topology. With that stuff out of the way...
There are multiple ways to compute $zeta(-1)$. You've outlined using the functional equation to relate it to $zeta(2)$. But the more relevant method is in the Wikipedia article:
$$-3zeta(-1)=eta(-1)=lim_{xto 1^-}left(1-2x+3x^2-4x^3+cdotsright)=lim_{xto 1^-}frac{1}{(1+x)^2}=frac14$$
where $eta$ is the Dirichlet eta function, not to be confused with the unrelated and better-known Dedekind eta function.
This is a direct analogue to what Ramanujan wrote in his notebook:
$$-3c=1-2+3-4+mathrm{&c}=frac{1}{(1+1)^2}=frac14$$
Imagine, for the sake of argument, that Ramanujan's intention is in fact to compute $zeta(-1)$. Then every single step in his computation of "$c$" corresponds to a step in the computation of $zeta(-1)$, just with the variables $s$ and $x$ elided. And the computation of $zeta(-1)$ is rigorous; each step is justified in the Wikipedia article, which I'll quote for completeness:
Where both Dirichlet series converge, one has the identities:
$$
begin{alignat}{7}
zeta(s)&{}={}&1^{-s}+2^{-s}&&{}+3^{-s}+4^{-s}&&{}+5^{-s}+6^{-s}+cdots& \
2times2^{-s}zeta(s)&{}={}& 2times2^{-s}&& {}+2times4^{-s}&&{} +2times6^{-s}+cdots& \
left(1-2^{1-s}right)zeta(s)&{}={}&1^{-s}-2^{-s}&&{}+3^{-s}-4^{-s}&&{}+5^{-s}-6^{-s}+cdots&=eta(s) \
end{alignat}
$$
The identity $(1-2^{1-s})zeta(s)=eta(s)$ continues to hold when both functions are extended by analytic continuation to include values of $s$ for which the above series diverge. Substituting $s = −1$, one gets $−3zeta(−1) = eta(−1)$. Now, computing $eta(−1)$ is an easier task, as the eta function is equal to the Abel sum of its defining series, which is a one-sided limit.
And that explains why Ramanujan's manipulation yields the same value as $zeta(-1)$.
Perhaps the most obscure step in the argument is the usage of Abel summation to evaluate a Dirichlet series. The reference to Knopp justifies that step better than I would be able to. Beyond that, I'm not sure what else you're likely to object to. Feel free to ask about any particular step!
answered Dec 18 '18 at 7:35
Chris CulterChris Culter
20k43483
$endgroup$
Well, "Ramanujan summation" usually refers to something else entirely, so let's not use that phrase here. Alternate topologies on the real numbers are another red herring; they're not relevant to this discussion. We're not evaluating the limit of a sequence of partial sums in any topology. With that stuff out of the way...
There are multiple ways to compute $zeta(-1)$. You've outlined using the functional equation to relate it to $zeta(2)$. But the more relevant method is in the Wikipedia article:
$$-3zeta(-1)=eta(-1)=lim_{xto 1^-}left(1-2x+3x^2-4x^3+cdotsright)=lim_{xto 1^-}frac{1}{(1+x)^2}=frac14$$
where $eta$ is the Dirichlet eta function, not to be confused with the unrelated and better-known Dedekind eta function.
This is a direct analogue to what Ramanujan wrote in his notebook:
$$-3c=1-2+3-4+mathrm{&c}=frac{1}{(1+1)^2}=frac14$$
Imagine, for the sake of argument, that Ramanujan's intention is in fact to compute $zeta(-1)$. Then every single step in his computation of "$c$" corresponds to a step in the computation of $zeta(-1)$, just with the variables $s$ and $x$ elided. And the computation of $zeta(-1)$ is rigorous; each step is justified in the Wikipedia article, which I'll quote for completeness:
Where both Dirichlet series converge, one has the identities:
$$
begin{alignat}{7}
zeta(s)&{}={}&1^{-s}+2^{-s}&&{}+3^{-s}+4^{-s}&&{}+5^{-s}+6^{-s}+cdots& \
2times2^{-s}zeta(s)&{}={}& 2times2^{-s}&& {}+2times4^{-s}&&{} +2times6^{-s}+cdots& \
left(1-2^{1-s}right)zeta(s)&{}={}&1^{-s}-2^{-s}&&{}+3^{-s}-4^{-s}&&{}+5^{-s}-6^{-s}+cdots&=eta(s) \
end{alignat}
$$
The identity $(1-2^{1-s})zeta(s)=eta(s)$ continues to hold when both functions are extended by analytic continuation to include values of $s$ for which the above series diverge. Substituting $s = −1$, one gets $−3zeta(−1) = eta(−1)$. Now, computing $eta(−1)$ is an easier task, as the eta function is equal to the Abel sum of its defining series, which is a one-sided limit.
And that explains why Ramanujan's manipulation yields the same value as $zeta(-1)$.
Perhaps the most obscure step in the argument is the usage of Abel summation to evaluate a Dirichlet series. The reference to Knopp justifies that step better than I would be able to. Beyond that, I'm not sure what else you're likely to object to. Feel free to ask about any particular step!
answered Dec 18 '18 at 7:35
Chris CulterChris Culter
20k43483
answered Dec 18 '18 at 7:35
Chris CulterChris Culter
20k43483
answered Dec 18 '18 at 7:35
Chris CulterChris Culter
20k43483
answered Dec 18 '18 at 7:35
Chris CulterChris Culter
20k43483
20k43483
1
$begingroup$
That "the eta function is equal to the Abel sum of its defining series" (so you can replace $lim_{s to -k}eta(s)$ by $lim_{z to 1} sum_n (-1)^{n+1} n^k z^n$) is not obvious : it is a consequence of $sum_{n=1}^N (-1)^{n+1} = frac{1+(-1)^{N+1}}{2}$ that summing by parts $K > -Re(s)$ times makes $sum_n (-1)^{n+1} n^{-s} z^n$ continuous in $|z| le 1, z ne -1$ and analytic in $s, Re(s) > -K$, whence $eta(s) =lim_{z to 1} sum_n (-1)^{n+1} n^{-s} z^n$ is valid for every $s$.
$endgroup$
– reuns
Dec 18 '18 at 7:43
$begingroup$
@reuns I do want to recognize that the equality isn't obvious! But it isn't an accident, either. The last time I read the Knopp reference, I got the impression that this phenomenon is not peculiar to this series. Rather, as a general principle, Abel summation works as well as you could possibly hope that it might for Dirichlet series. And there is some rigorous formulation of that last sentence. Am I off base there?
$endgroup$
– Chris Culter
Dec 18 '18 at 7:53
$begingroup$
It doesn't work with $sum_n n^{-s}$, only with Dirichlet series that summing by parts yields an analytic continuation
$endgroup$
– reuns
Dec 18 '18 at 8:02
$begingroup$
@reuns Sure, that series is too much to hope for. I meant "work" in the more lax sense of, where the Abel sum does exist, it agrees with the analytic continuation. That's all we need.
$endgroup$
– Chris Culter
Dec 18 '18 at 8:07
1
$begingroup$
Let $f(t) = sum_{n=1}^infty a_n e^{-nt}$ with $a_n = O(n^k)$ and $F(s) = sum_{n=1}^infty a_n n^{-s}, Re(s) > k+1$. Then $int_0^infty t^{s-1}f(t)dt = Gamma(s) F(s)$ and if $lim_{t to 0^+} f(t) = C ne 0$ converges then $int_0^infty t^{s-1}f(t)dt$ converges and is analytic for $Re(s) > 0$, obtaining that (the continuation of) $F(s)$ is analytic and $lim_{s to 0^+} F(s)=C$. But summing by parts $sum_{n=1}^infty a_n n^{-s}$ would have given the same result.
$endgroup$
– reuns
Dec 18 '18 at 8:40
add a comment |
1
$begingroup$
That "the eta function is equal to the Abel sum of its defining series" (so you can replace $lim_{s to -k}eta(s)$ by $lim_{z to 1} sum_n (-1)^{n+1} n^k z^n$) is not obvious : it is a consequence of $sum_{n=1}^N (-1)^{n+1} = frac{1+(-1)^{N+1}}{2}$ that summing by parts $K > -Re(s)$ times makes $sum_n (-1)^{n+1} n^{-s} z^n$ continuous in $|z| le 1, z ne -1$ and analytic in $s, Re(s) > -K$, whence $eta(s) =lim_{z to 1} sum_n (-1)^{n+1} n^{-s} z^n$ is valid for every $s$.
$endgroup$
– reuns
Dec 18 '18 at 7:43
$begingroup$
@reuns I do want to recognize that the equality isn't obvious! But it isn't an accident, either. The last time I read the Knopp reference, I got the impression that this phenomenon is not peculiar to this series. Rather, as a general principle, Abel summation works as well as you could possibly hope that it might for Dirichlet series. And there is some rigorous formulation of that last sentence. Am I off base there?
$endgroup$
– Chris Culter
Dec 18 '18 at 7:53
$begingroup$
It doesn't work with $sum_n n^{-s}$, only with Dirichlet series that summing by parts yields an analytic continuation
$endgroup$
– reuns
Dec 18 '18 at 8:02
$begingroup$
@reuns Sure, that series is too much to hope for. I meant "work" in the more lax sense of, where the Abel sum does exist, it agrees with the analytic continuation. That's all we need.
$endgroup$
– Chris Culter
Dec 18 '18 at 8:07
1
$begingroup$
Let $f(t) = sum_{n=1}^infty a_n e^{-nt}$ with $a_n = O(n^k)$ and $F(s) = sum_{n=1}^infty a_n n^{-s}, Re(s) > k+1$. Then $int_0^infty t^{s-1}f(t)dt = Gamma(s) F(s)$ and if $lim_{t to 0^+} f(t) = C ne 0$ converges then $int_0^infty t^{s-1}f(t)dt$ converges and is analytic for $Re(s) > 0$, obtaining that (the continuation of) $F(s)$ is analytic and $lim_{s to 0^+} F(s)=C$. But summing by parts $sum_{n=1}^infty a_n n^{-s}$ would have given the same result.
$endgroup$
– reuns
Dec 18 '18 at 8:40
1
1
$begingroup$
That "the eta function is equal to the Abel sum of its defining series" (so you can replace $lim_{s to -k}eta(s)$ by $lim_{z to 1} sum_n (-1)^{n+1} n^k z^n$) is not obvious : it is a consequence of $sum_{n=1}^N (-1)^{n+1} = frac{1+(-1)^{N+1}}{2}$ that summing by parts $K > -Re(s)$ times makes $sum_n (-1)^{n+1} n^{-s} z^n$ continuous in $|z| le 1, z ne -1$ and analytic in $s, Re(s) > -K$, whence $eta(s) =lim_{z to 1} sum_n (-1)^{n+1} n^{-s} z^n$ is valid for every $s$.
$endgroup$
– reuns
Dec 18 '18 at 7:43
$begingroup$
That "the eta function is equal to the Abel sum of its defining series" (so you can replace $lim_{s to -k}eta(s)$ by $lim_{z to 1} sum_n (-1)^{n+1} n^k z^n$) is not obvious : it is a consequence of $sum_{n=1}^N (-1)^{n+1} = frac{1+(-1)^{N+1}}{2}$ that summing by parts $K > -Re(s)$ times makes $sum_n (-1)^{n+1} n^{-s} z^n$ continuous in $|z| le 1, z ne -1$ and analytic in $s, Re(s) > -K$, whence $eta(s) =lim_{z to 1} sum_n (-1)^{n+1} n^{-s} z^n$ is valid for every $s$.
$endgroup$
– reuns
Dec 18 '18 at 7:43
$begingroup$
@reuns I do want to recognize that the equality isn't obvious! But it isn't an accident, either. The last time I read the Knopp reference, I got the impression that this phenomenon is not peculiar to this series. Rather, as a general principle, Abel summation works as well as you could possibly hope that it might for Dirichlet series. And there is some rigorous formulation of that last sentence. Am I off base there?
$endgroup$
– Chris Culter
Dec 18 '18 at 7:53
$begingroup$
@reuns I do want to recognize that the equality isn't obvious! But it isn't an accident, either. The last time I read the Knopp reference, I got the impression that this phenomenon is not peculiar to this series. Rather, as a general principle, Abel summation works as well as you could possibly hope that it might for Dirichlet series. And there is some rigorous formulation of that last sentence. Am I off base there?
$endgroup$
– Chris Culter
Dec 18 '18 at 7:53
$begingroup$
It doesn't work with $sum_n n^{-s}$, only with Dirichlet series that summing by parts yields an analytic continuation
$endgroup$
– reuns
Dec 18 '18 at 8:02
$begingroup$
It doesn't work with $sum_n n^{-s}$, only with Dirichlet series that summing by parts yields an analytic continuation
$endgroup$
– reuns
Dec 18 '18 at 8:02
$begingroup$
@reuns Sure, that series is too much to hope for. I meant "work" in the more lax sense of, where the Abel sum does exist, it agrees with the analytic continuation. That's all we need.
$endgroup$
– Chris Culter
Dec 18 '18 at 8:07
$begingroup$
@reuns Sure, that series is too much to hope for. I meant "work" in the more lax sense of, where the Abel sum does exist, it agrees with the analytic continuation. That's all we need.
$endgroup$
– Chris Culter
Dec 18 '18 at 8:07
1
1
$begingroup$
Let $f(t) = sum_{n=1}^infty a_n e^{-nt}$ with $a_n = O(n^k)$ and $F(s) = sum_{n=1}^infty a_n n^{-s}, Re(s) > k+1$. Then $int_0^infty t^{s-1}f(t)dt = Gamma(s) F(s)$ and if $lim_{t to 0^+} f(t) = C ne 0$ converges then $int_0^infty t^{s-1}f(t)dt$ converges and is analytic for $Re(s) > 0$, obtaining that (the continuation of) $F(s)$ is analytic and $lim_{s to 0^+} F(s)=C$. But summing by parts $sum_{n=1}^infty a_n n^{-s}$ would have given the same result.
$endgroup$
– reuns
Dec 18 '18 at 8:40
$begingroup$
Let $f(t) = sum_{n=1}^infty a_n e^{-nt}$ with $a_n = O(n^k)$ and $F(s) = sum_{n=1}^infty a_n n^{-s}, Re(s) > k+1$. Then $int_0^infty t^{s-1}f(t)dt = Gamma(s) F(s)$ and if $lim_{t to 0^+} f(t) = C ne 0$ converges then $int_0^infty t^{s-1}f(t)dt$ converges and is analytic for $Re(s) > 0$, obtaining that (the continuation of) $F(s)$ is analytic and $lim_{s to 0^+} F(s)=C$. But summing by parts $sum_{n=1}^infty a_n n^{-s}$ would have given the same result.
$endgroup$
– reuns
Dec 18 '18 at 8:40
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044810%2fis-there-a-connection-between-zeta-1-and-ramanujans-calculation-of-the-sum%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Nitpick: $zeta(s) = 2^s pi^{s-1} sinleft(frac{pi s}{2} right) Gamma (1-s) zeta(1-s)$ only gives $zeta(s)$ for $operatorname{Re}s<0$ if you start with the standard series. For the critical strip you need more.
$endgroup$
– Arthur
Dec 18 '18 at 7:33
$begingroup$
This "equality" occurs over and over again. At least in this question, it is highlighted that it does not hold in the usual sense
$endgroup$
– Peter
Dec 22 '18 at 14:38