The divergence of the series of reciprocals of primes (proof check):
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I want to check my attempt at a proof for the divergence of
$$sum_{n=1}^{infty} frac{1}{p_n} tag{ $star$ }.$$
We begin with assuming that $(star)$ converges. If $(star)$ converges, there is an integer $a$ so that, $$sum_{j=a+1}^{infty}frac{1}{p_j} lt frac{1}{b}$$ where $b>1$. Note that given any $b$ there exists an $a$ that satisfies the above inequality. Now, we let $M = p_1cdotcdotcdot p_a$ and consider the number $1+ nM$ for $n = 1,2,dots$ Any factors of $1+nM$ are $p_i$ for $i geq a+1,a+2,dots$ Hence, we can write for each $g geq 1$:
$$sum_{n=1}^{g} frac{1}{1+nM} leq sum_{x=1}^{infty}(sum_{j=a+1}^{infty}frac{1}{p_j})^x$$
But on the right hand side, we have a geometric series:
$$sum_{n=1}^{g} frac{1}{1+nM} leq sum_{x=1}^{infty}(frac{1}{b})^x$$
Since the geometric series converges, it means that $sum_{n=1}^{infty} frac{1}{1+nM}$ converges and is bounded above. But that is a contradiction because if we do the integral test on $sum_{n=1}^{infty} frac{1}{1+nM}$, it diverges. Therefore, $(star)$ diverges.
number-theory prime-numbers divergent-series
edited Dec 1 at 4:08
Klangen
1,25811129
asked May 20 '13 at 3:49
kvmu
317211
|
show 2 more comments
up vote
6
down vote
favorite
I want to check my attempt at a proof for the divergence of
$$sum_{n=1}^{infty} frac{1}{p_n} tag{ $star$ }.$$
We begin with assuming that $(star)$ converges. If $(star)$ converges, there is an integer $a$ so that, $$sum_{j=a+1}^{infty}frac{1}{p_j} lt frac{1}{b}$$ where $b>1$. Note that given any $b$ there exists an $a$ that satisfies the above inequality. Now, we let $M = p_1cdotcdotcdot p_a$ and consider the number $1+ nM$ for $n = 1,2,dots$ Any factors of $1+nM$ are $p_i$ for $i geq a+1,a+2,dots$ Hence, we can write for each $g geq 1$:
$$sum_{n=1}^{g} frac{1}{1+nM} leq sum_{x=1}^{infty}(sum_{j=a+1}^{infty}frac{1}{p_j})^x$$
But on the right hand side, we have a geometric series:
$$sum_{n=1}^{g} frac{1}{1+nM} leq sum_{x=1}^{infty}(frac{1}{b})^x$$
Since the geometric series converges, it means that $sum_{n=1}^{infty} frac{1}{1+nM}$ converges and is bounded above. But that is a contradiction because if we do the integral test on $sum_{n=1}^{infty} frac{1}{1+nM}$, it diverges. Therefore, $(star)$ diverges.
number-theory prime-numbers divergent-series
edited Dec 1 at 4:08
Klangen
1,25811129
asked May 20 '13 at 3:49
kvmu
317211
Alright! So, from the definition of M, we can say: $frac{1}{1+nM} = frac{1}{p_{k+1}p_{k+2}cdots p_{x}}$, so this term must appear somewhere in the expansion of $(frac{1}{p_{k+1}}+frac{1}{p_{k+2}}+frac{1}{p_{k+3}}+cdots)^x$ - that's how that inequality came to be.
– kvmu
May 20 '13 at 4:02
1
Your statement "Note that as b changes, a changes correspondingly.", though correct, would be better expressed as "given any b, there is an a such that:". This can be valuable in infinite descent proofs, where $a lt b$,then $a$ satisfies the criterion, so you can apply it again...
– Ross Millikan
May 20 '13 at 4:36
Oh, good call - I will edit that right now, but other than that is the proof okay?
– kvmu
May 20 '13 at 4:43
2
Some nitpicks: Clearly you need $b>1$ and you probably meant $igeq a+1$
– Alex R.
May 20 '13 at 5:03
You might wanna compare your proof to the proof on pg. 18-19 of Tom Apostol's text "Introduction to Analytic Number Theory". I could provide both pages if you like..
– user70962
May 20 '13 at 7:51
|
show 2 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I want to check my attempt at a proof for the divergence of
$$sum_{n=1}^{infty} frac{1}{p_n} tag{ $star$ }.$$
We begin with assuming that $(star)$ converges. If $(star)$ converges, there is an integer $a$ so that, $$sum_{j=a+1}^{infty}frac{1}{p_j} lt frac{1}{b}$$ where $b>1$. Note that given any $b$ there exists an $a$ that satisfies the above inequality. Now, we let $M = p_1cdotcdotcdot p_a$ and consider the number $1+ nM$ for $n = 1,2,dots$ Any factors of $1+nM$ are $p_i$ for $i geq a+1,a+2,dots$ Hence, we can write for each $g geq 1$:
$$sum_{n=1}^{g} frac{1}{1+nM} leq sum_{x=1}^{infty}(sum_{j=a+1}^{infty}frac{1}{p_j})^x$$
But on the right hand side, we have a geometric series:
$$sum_{n=1}^{g} frac{1}{1+nM} leq sum_{x=1}^{infty}(frac{1}{b})^x$$
Since the geometric series converges, it means that $sum_{n=1}^{infty} frac{1}{1+nM}$ converges and is bounded above. But that is a contradiction because if we do the integral test on $sum_{n=1}^{infty} frac{1}{1+nM}$, it diverges. Therefore, $(star)$ diverges.
number-theory prime-numbers divergent-series
edited Dec 1 at 4:08
Klangen
1,25811129
asked May 20 '13 at 3:49
kvmu
317211
I want to check my attempt at a proof for the divergence of
$$sum_{n=1}^{infty} frac{1}{p_n} tag{ $star$ }.$$
We begin with assuming that $(star)$ converges. If $(star)$ converges, there is an integer $a$ so that, $$sum_{j=a+1}^{infty}frac{1}{p_j} lt frac{1}{b}$$ where $b>1$. Note that given any $b$ there exists an $a$ that satisfies the above inequality. Now, we let $M = p_1cdotcdotcdot p_a$ and consider the number $1+ nM$ for $n = 1,2,dots$ Any factors of $1+nM$ are $p_i$ for $i geq a+1,a+2,dots$ Hence, we can write for each $g geq 1$:
$$sum_{n=1}^{g} frac{1}{1+nM} leq sum_{x=1}^{infty}(sum_{j=a+1}^{infty}frac{1}{p_j})^x$$
But on the right hand side, we have a geometric series:
$$sum_{n=1}^{g} frac{1}{1+nM} leq sum_{x=1}^{infty}(frac{1}{b})^x$$
Since the geometric series converges, it means that $sum_{n=1}^{infty} frac{1}{1+nM}$ converges and is bounded above. But that is a contradiction because if we do the integral test on $sum_{n=1}^{infty} frac{1}{1+nM}$, it diverges. Therefore, $(star)$ diverges.
number-theory prime-numbers divergent-series
number-theory prime-numbers divergent-series
edited Dec 1 at 4:08
Klangen
1,25811129
asked May 20 '13 at 3:49
kvmu
317211
edited Dec 1 at 4:08
Klangen
1,25811129
asked May 20 '13 at 3:49
kvmu
317211
edited Dec 1 at 4:08
Klangen
1,25811129
edited Dec 1 at 4:08
Klangen
1,25811129
edited Dec 1 at 4:08
Klangen
1,25811129
1,25811129
asked May 20 '13 at 3:49
kvmu
317211
asked May 20 '13 at 3:49
kvmu
317211
asked May 20 '13 at 3:49
kvmu
317211
317211
Alright! So, from the definition of M, we can say: $frac{1}{1+nM} = frac{1}{p_{k+1}p_{k+2}cdots p_{x}}$, so this term must appear somewhere in the expansion of $(frac{1}{p_{k+1}}+frac{1}{p_{k+2}}+frac{1}{p_{k+3}}+cdots)^x$ - that's how that inequality came to be.
– kvmu
May 20 '13 at 4:02
1
Your statement "Note that as b changes, a changes correspondingly.", though correct, would be better expressed as "given any b, there is an a such that:". This can be valuable in infinite descent proofs, where $a lt b$,then $a$ satisfies the criterion, so you can apply it again...
– Ross Millikan
May 20 '13 at 4:36
Oh, good call - I will edit that right now, but other than that is the proof okay?
– kvmu
May 20 '13 at 4:43
2
Some nitpicks: Clearly you need $b>1$ and you probably meant $igeq a+1$
– Alex R.
May 20 '13 at 5:03
You might wanna compare your proof to the proof on pg. 18-19 of Tom Apostol's text "Introduction to Analytic Number Theory". I could provide both pages if you like..
– user70962
May 20 '13 at 7:51
|
show 2 more comments
Alright! So, from the definition of M, we can say: $frac{1}{1+nM} = frac{1}{p_{k+1}p_{k+2}cdots p_{x}}$, so this term must appear somewhere in the expansion of $(frac{1}{p_{k+1}}+frac{1}{p_{k+2}}+frac{1}{p_{k+3}}+cdots)^x$ - that's how that inequality came to be.
– kvmu
May 20 '13 at 4:02
1
Your statement "Note that as b changes, a changes correspondingly.", though correct, would be better expressed as "given any b, there is an a such that:". This can be valuable in infinite descent proofs, where $a lt b$,then $a$ satisfies the criterion, so you can apply it again...
– Ross Millikan
May 20 '13 at 4:36
Oh, good call - I will edit that right now, but other than that is the proof okay?
– kvmu
May 20 '13 at 4:43
2
Some nitpicks: Clearly you need $b>1$ and you probably meant $igeq a+1$
– Alex R.
May 20 '13 at 5:03
You might wanna compare your proof to the proof on pg. 18-19 of Tom Apostol's text "Introduction to Analytic Number Theory". I could provide both pages if you like..
– user70962
May 20 '13 at 7:51
Alright! So, from the definition of M, we can say: $frac{1}{1+nM} = frac{1}{p_{k+1}p_{k+2}cdots p_{x}}$, so this term must appear somewhere in the expansion of $(frac{1}{p_{k+1}}+frac{1}{p_{k+2}}+frac{1}{p_{k+3}}+cdots)^x$ - that's how that inequality came to be.
– kvmu
May 20 '13 at 4:02
Alright! So, from the definition of M, we can say: $frac{1}{1+nM} = frac{1}{p_{k+1}p_{k+2}cdots p_{x}}$, so this term must appear somewhere in the expansion of $(frac{1}{p_{k+1}}+frac{1}{p_{k+2}}+frac{1}{p_{k+3}}+cdots)^x$ - that's how that inequality came to be.
– kvmu
May 20 '13 at 4:02
1
1
Your statement "Note that as b changes, a changes correspondingly.", though correct, would be better expressed as "given any b, there is an a such that:". This can be valuable in infinite descent proofs, where $a lt b$,then $a$ satisfies the criterion, so you can apply it again...
– Ross Millikan
May 20 '13 at 4:36
Your statement "Note that as b changes, a changes correspondingly.", though correct, would be better expressed as "given any b, there is an a such that:". This can be valuable in infinite descent proofs, where $a lt b$,then $a$ satisfies the criterion, so you can apply it again...
– Ross Millikan
May 20 '13 at 4:36
Oh, good call - I will edit that right now, but other than that is the proof okay?
– kvmu
May 20 '13 at 4:43
Oh, good call - I will edit that right now, but other than that is the proof okay?
– kvmu
May 20 '13 at 4:43
2
2
Some nitpicks: Clearly you need $b>1$ and you probably meant $igeq a+1$
– Alex R.
May 20 '13 at 5:03
Some nitpicks: Clearly you need $b>1$ and you probably meant $igeq a+1$
– Alex R.
May 20 '13 at 5:03
You might wanna compare your proof to the proof on pg. 18-19 of Tom Apostol's text "Introduction to Analytic Number Theory". I could provide both pages if you like..
– user70962
May 20 '13 at 7:51
You might wanna compare your proof to the proof on pg. 18-19 of Tom Apostol's text "Introduction to Analytic Number Theory". I could provide both pages if you like..
– user70962
May 20 '13 at 7:51
|
show 2 more comments
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Alright! So, from the definition of M, we can say: $frac{1}{1+nM} = frac{1}{p_{k+1}p_{k+2}cdots p_{x}}$, so this term must appear somewhere in the expansion of $(frac{1}{p_{k+1}}+frac{1}{p_{k+2}}+frac{1}{p_{k+3}}+cdots)^x$ - that's how that inequality came to be.
– kvmu
May 20 '13 at 4:02
1
Your statement "Note that as b changes, a changes correspondingly.", though correct, would be better expressed as "given any b, there is an a such that:". This can be valuable in infinite descent proofs, where $a lt b$,then $a$ satisfies the criterion, so you can apply it again...
– Ross Millikan
May 20 '13 at 4:36
Oh, good call - I will edit that right now, but other than that is the proof okay?
– kvmu
May 20 '13 at 4:43
2
Some nitpicks: Clearly you need $b>1$ and you probably meant $igeq a+1$
– Alex R.
May 20 '13 at 5:03
You might wanna compare your proof to the proof on pg. 18-19 of Tom Apostol's text "Introduction to Analytic Number Theory". I could provide both pages if you like..
– user70962
May 20 '13 at 7:51