The partial sum and partial product of $zeta$function
$begingroup$
Taking the partial sum of the $zeta $ function:
$$zeta^H(s,k)=sum_{n=0}^k frac{1}{n^s}$$
and the partial product f the $zeta $ function:
$$zeta^P(s,j)=prod_{i=0}^j frac{1}{1-p_i^{-s}}$$
I have two questions:
(1) For $p$ being prime and $n,k,iinBbb N$; do there exist closed functions for $zeta^H(s,k)$ and $zeta^P(s,j)$ to which the partial sum respectively the partial product converge?
(2) Does anybody know from a proof that would confirm that for a given $s$, if and if only $j,ktoinfty$ then:
$$zeta^H(s,k)-zeta^P(s,j)to0$$
Thanks for your help in advance.
riemann-zeta euler-product
asked May 20 '13 at 19:00
al-Hwarizmial-Hwarizmi
2,64511131
$endgroup$
add a comment |
$begingroup$
Taking the partial sum of the $zeta $ function:
$$zeta^H(s,k)=sum_{n=0}^k frac{1}{n^s}$$
and the partial product f the $zeta $ function:
$$zeta^P(s,j)=prod_{i=0}^j frac{1}{1-p_i^{-s}}$$
I have two questions:
(1) For $p$ being prime and $n,k,iinBbb N$; do there exist closed functions for $zeta^H(s,k)$ and $zeta^P(s,j)$ to which the partial sum respectively the partial product converge?
(2) Does anybody know from a proof that would confirm that for a given $s$, if and if only $j,ktoinfty$ then:
$$zeta^H(s,k)-zeta^P(s,j)to0$$
Thanks for your help in advance.
riemann-zeta euler-product
asked May 20 '13 at 19:00
al-Hwarizmial-Hwarizmi
2,64511131
$endgroup$
$begingroup$
If the real part of $s$ is greater than 1, then both finite product and finite sum converge absolutely to $zeta(s)$.
$endgroup$
– i707107
May 20 '13 at 20:04
$begingroup$
sure, this is trivial. But my questions were about what happens before?
$endgroup$
– al-Hwarizmi
May 21 '13 at 11:45
add a comment |
$begingroup$
Taking the partial sum of the $zeta $ function:
$$zeta^H(s,k)=sum_{n=0}^k frac{1}{n^s}$$
and the partial product f the $zeta $ function:
$$zeta^P(s,j)=prod_{i=0}^j frac{1}{1-p_i^{-s}}$$
I have two questions:
(1) For $p$ being prime and $n,k,iinBbb N$; do there exist closed functions for $zeta^H(s,k)$ and $zeta^P(s,j)$ to which the partial sum respectively the partial product converge?
(2) Does anybody know from a proof that would confirm that for a given $s$, if and if only $j,ktoinfty$ then:
$$zeta^H(s,k)-zeta^P(s,j)to0$$
Thanks for your help in advance.
riemann-zeta euler-product
asked May 20 '13 at 19:00
al-Hwarizmial-Hwarizmi
2,64511131
$endgroup$
Taking the partial sum of the $zeta $ function:
$$zeta^H(s,k)=sum_{n=0}^k frac{1}{n^s}$$
and the partial product f the $zeta $ function:
$$zeta^P(s,j)=prod_{i=0}^j frac{1}{1-p_i^{-s}}$$
I have two questions:
(1) For $p$ being prime and $n,k,iinBbb N$; do there exist closed functions for $zeta^H(s,k)$ and $zeta^P(s,j)$ to which the partial sum respectively the partial product converge?
(2) Does anybody know from a proof that would confirm that for a given $s$, if and if only $j,ktoinfty$ then:
$$zeta^H(s,k)-zeta^P(s,j)to0$$
Thanks for your help in advance.
riemann-zeta euler-product
riemann-zeta euler-product
asked May 20 '13 at 19:00
al-Hwarizmial-Hwarizmi
2,64511131
asked May 20 '13 at 19:00
al-Hwarizmial-Hwarizmi
2,64511131
asked May 20 '13 at 19:00
al-Hwarizmial-Hwarizmi
2,64511131
asked May 20 '13 at 19:00
al-Hwarizmial-Hwarizmi
2,64511131
asked May 20 '13 at 19:00
al-Hwarizmial-Hwarizmi
2,64511131
2,64511131
$begingroup$
If the real part of $s$ is greater than 1, then both finite product and finite sum converge absolutely to $zeta(s)$.
$endgroup$
– i707107
May 20 '13 at 20:04
$begingroup$
sure, this is trivial. But my questions were about what happens before?
$endgroup$
– al-Hwarizmi
May 21 '13 at 11:45
add a comment |
$begingroup$
If the real part of $s$ is greater than 1, then both finite product and finite sum converge absolutely to $zeta(s)$.
$endgroup$
– i707107
May 20 '13 at 20:04
$begingroup$
sure, this is trivial. But my questions were about what happens before?
$endgroup$
– al-Hwarizmi
May 21 '13 at 11:45
$begingroup$
If the real part of $s$ is greater than 1, then both finite product and finite sum converge absolutely to $zeta(s)$.
$endgroup$
– i707107
May 20 '13 at 20:04
$begingroup$
If the real part of $s$ is greater than 1, then both finite product and finite sum converge absolutely to $zeta(s)$.
$endgroup$
– i707107
May 20 '13 at 20:04
$begingroup$
sure, this is trivial. But my questions were about what happens before?
$endgroup$
– al-Hwarizmi
May 21 '13 at 11:45
$begingroup$
sure, this is trivial. But my questions were about what happens before?
$endgroup$
– al-Hwarizmi
May 21 '13 at 11:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $s=1$ and positive nonzero integer $k$ you have the harmonic numbers. Their numerators are in the OEIS as sequence A001008 and their denominators as A002805.
That being said, there are other formulas that express your function $zeta^H$. For instance, this paper by Borwein, Fee, Ferguson and Van Der Waall proposes the following (by denoting your function $zeta^H (s,k)$ by $zeta_N(s)=displaystylesum_{n=1}^{N}n^{-s}$):
$$
zeta(s) = zeta_{N-1}(s) +N^{-s} + frac{N^{1-s}}{a-1} + sum_{n=1}^{infty} frac{B_{2n}}{(2n)!}left( prod_{j=0}^{2n-2} (s+j) right) N^{1-s-2n} + R,
$$
where the $B_n$ are the Bernouilli numbers.
answered Dec 17 '18 at 11:57
KlangenKlangen
1,71011334
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For $s=1$ and positive nonzero integer $k$ you have the harmonic numbers. Their numerators are in the OEIS as sequence A001008 and their denominators as A002805.
That being said, there are other formulas that express your function $zeta^H$. For instance, this paper by Borwein, Fee, Ferguson and Van Der Waall proposes the following (by denoting your function $zeta^H (s,k)$ by $zeta_N(s)=displaystylesum_{n=1}^{N}n^{-s}$):
$$
zeta(s) = zeta_{N-1}(s) +N^{-s} + frac{N^{1-s}}{a-1} + sum_{n=1}^{infty} frac{B_{2n}}{(2n)!}left( prod_{j=0}^{2n-2} (s+j) right) N^{1-s-2n} + R,
$$
where the $B_n$ are the Bernouilli numbers.
answered Dec 17 '18 at 11:57
KlangenKlangen
1,71011334
$endgroup$
add a comment |
$begingroup$
For $s=1$ and positive nonzero integer $k$ you have the harmonic numbers. Their numerators are in the OEIS as sequence A001008 and their denominators as A002805.
That being said, there are other formulas that express your function $zeta^H$. For instance, this paper by Borwein, Fee, Ferguson and Van Der Waall proposes the following (by denoting your function $zeta^H (s,k)$ by $zeta_N(s)=displaystylesum_{n=1}^{N}n^{-s}$):
$$
zeta(s) = zeta_{N-1}(s) +N^{-s} + frac{N^{1-s}}{a-1} + sum_{n=1}^{infty} frac{B_{2n}}{(2n)!}left( prod_{j=0}^{2n-2} (s+j) right) N^{1-s-2n} + R,
$$
where the $B_n$ are the Bernouilli numbers.
answered Dec 17 '18 at 11:57
KlangenKlangen
1,71011334
$endgroup$
add a comment |
$begingroup$
For $s=1$ and positive nonzero integer $k$ you have the harmonic numbers. Their numerators are in the OEIS as sequence A001008 and their denominators as A002805.
That being said, there are other formulas that express your function $zeta^H$. For instance, this paper by Borwein, Fee, Ferguson and Van Der Waall proposes the following (by denoting your function $zeta^H (s,k)$ by $zeta_N(s)=displaystylesum_{n=1}^{N}n^{-s}$):
$$
zeta(s) = zeta_{N-1}(s) +N^{-s} + frac{N^{1-s}}{a-1} + sum_{n=1}^{infty} frac{B_{2n}}{(2n)!}left( prod_{j=0}^{2n-2} (s+j) right) N^{1-s-2n} + R,
$$
where the $B_n$ are the Bernouilli numbers.
answered Dec 17 '18 at 11:57
KlangenKlangen
1,71011334
$endgroup$
For $s=1$ and positive nonzero integer $k$ you have the harmonic numbers. Their numerators are in the OEIS as sequence A001008 and their denominators as A002805.
That being said, there are other formulas that express your function $zeta^H$. For instance, this paper by Borwein, Fee, Ferguson and Van Der Waall proposes the following (by denoting your function $zeta^H (s,k)$ by $zeta_N(s)=displaystylesum_{n=1}^{N}n^{-s}$):
$$
zeta(s) = zeta_{N-1}(s) +N^{-s} + frac{N^{1-s}}{a-1} + sum_{n=1}^{infty} frac{B_{2n}}{(2n)!}left( prod_{j=0}^{2n-2} (s+j) right) N^{1-s-2n} + R,
$$
where the $B_n$ are the Bernouilli numbers.
answered Dec 17 '18 at 11:57
KlangenKlangen
1,71011334
answered Dec 17 '18 at 11:57
KlangenKlangen
1,71011334
answered Dec 17 '18 at 11:57
KlangenKlangen
1,71011334
answered Dec 17 '18 at 11:57
KlangenKlangen
1,71011334
1,71011334
add a comment |
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$begingroup$
If the real part of $s$ is greater than 1, then both finite product and finite sum converge absolutely to $zeta(s)$.
$endgroup$
– i707107
May 20 '13 at 20:04
$begingroup$
sure, this is trivial. But my questions were about what happens before?
$endgroup$
– al-Hwarizmi
May 21 '13 at 11:45