Does this series converge to a rational multiple of $pi^2$?
$begingroup$
If we define the sum
$$S_2=sum_{n_2=1}^inftyfrac{1}{(n_2)^2} $$
in which $n_2$ are products of an even number of prime factors, together with 1, so $n_2=1,4,6,...,15,16,21,22,...,24,$ etc., the sum seems to be a rational multiple of $pi^2$. Of course it converges by comparison to $zeta(2).$
The first $k=9,997,745$ of these (including 1) sum to approximately $frac{7pi^2}{60}$ and the ratio $S_k/(7pi^2/60)=0.999999978...$ (there are 7 nines in this decimal).
My unscientific test is that if Mathematica rounds to one it bears a second look. In this case I looked at various proofs of the Basel problem and connections with 2-primes $pq$ to see if there was a way of showing this.
Is this already known to be true? If not, can someone show it? Also any comments on the likelihood of the relation given the approximation are welcome. How much numerical similarity is enough to guess in this way?
I am still working on this, and if I find a proof or a reference I will post it.
sequences-and-series
asked Dec 6 '18 at 14:32
danieldaniel
6,25722259
$endgroup$
add a comment |
$begingroup$
If we define the sum
$$S_2=sum_{n_2=1}^inftyfrac{1}{(n_2)^2} $$
in which $n_2$ are products of an even number of prime factors, together with 1, so $n_2=1,4,6,...,15,16,21,22,...,24,$ etc., the sum seems to be a rational multiple of $pi^2$. Of course it converges by comparison to $zeta(2).$
The first $k=9,997,745$ of these (including 1) sum to approximately $frac{7pi^2}{60}$ and the ratio $S_k/(7pi^2/60)=0.999999978...$ (there are 7 nines in this decimal).
My unscientific test is that if Mathematica rounds to one it bears a second look. In this case I looked at various proofs of the Basel problem and connections with 2-primes $pq$ to see if there was a way of showing this.
Is this already known to be true? If not, can someone show it? Also any comments on the likelihood of the relation given the approximation are welcome. How much numerical similarity is enough to guess in this way?
I am still working on this, and if I find a proof or a reference I will post it.
sequences-and-series
asked Dec 6 '18 at 14:32
danieldaniel
6,25722259
$endgroup$
1
$begingroup$
A side note: $1 + zeta(2) P(2) = sum_{n=1}^{infty} frac{omega(n)}{n^2}$, where $omega(n)$ is the number of distinct prime divisors of $n$ and $P$ is the prime zeta function. Although you are interested in $sum_{n = 2|omega(n)}^{infty}frac{1}{n^2}$ I wonder if you can utilise some kind of mobius inversion or if you would just go in circles?
$endgroup$
– DanielOnMSE
Dec 6 '18 at 14:59
add a comment |
$begingroup$
If we define the sum
$$S_2=sum_{n_2=1}^inftyfrac{1}{(n_2)^2} $$
in which $n_2$ are products of an even number of prime factors, together with 1, so $n_2=1,4,6,...,15,16,21,22,...,24,$ etc., the sum seems to be a rational multiple of $pi^2$. Of course it converges by comparison to $zeta(2).$
The first $k=9,997,745$ of these (including 1) sum to approximately $frac{7pi^2}{60}$ and the ratio $S_k/(7pi^2/60)=0.999999978...$ (there are 7 nines in this decimal).
My unscientific test is that if Mathematica rounds to one it bears a second look. In this case I looked at various proofs of the Basel problem and connections with 2-primes $pq$ to see if there was a way of showing this.
Is this already known to be true? If not, can someone show it? Also any comments on the likelihood of the relation given the approximation are welcome. How much numerical similarity is enough to guess in this way?
I am still working on this, and if I find a proof or a reference I will post it.
sequences-and-series
asked Dec 6 '18 at 14:32
danieldaniel
6,25722259
$endgroup$
If we define the sum
$$S_2=sum_{n_2=1}^inftyfrac{1}{(n_2)^2} $$
in which $n_2$ are products of an even number of prime factors, together with 1, so $n_2=1,4,6,...,15,16,21,22,...,24,$ etc., the sum seems to be a rational multiple of $pi^2$. Of course it converges by comparison to $zeta(2).$
The first $k=9,997,745$ of these (including 1) sum to approximately $frac{7pi^2}{60}$ and the ratio $S_k/(7pi^2/60)=0.999999978...$ (there are 7 nines in this decimal).
My unscientific test is that if Mathematica rounds to one it bears a second look. In this case I looked at various proofs of the Basel problem and connections with 2-primes $pq$ to see if there was a way of showing this.
Is this already known to be true? If not, can someone show it? Also any comments on the likelihood of the relation given the approximation are welcome. How much numerical similarity is enough to guess in this way?
I am still working on this, and if I find a proof or a reference I will post it.
sequences-and-series
sequences-and-series
asked Dec 6 '18 at 14:32
danieldaniel
6,25722259
asked Dec 6 '18 at 14:32
danieldaniel
6,25722259
asked Dec 6 '18 at 14:32
danieldaniel
6,25722259
asked Dec 6 '18 at 14:32
danieldaniel
6,25722259
asked Dec 6 '18 at 14:32
danieldaniel
6,25722259
6,25722259
1
$begingroup$
A side note: $1 + zeta(2) P(2) = sum_{n=1}^{infty} frac{omega(n)}{n^2}$, where $omega(n)$ is the number of distinct prime divisors of $n$ and $P$ is the prime zeta function. Although you are interested in $sum_{n = 2|omega(n)}^{infty}frac{1}{n^2}$ I wonder if you can utilise some kind of mobius inversion or if you would just go in circles?
$endgroup$
– DanielOnMSE
Dec 6 '18 at 14:59
add a comment |
1
$begingroup$
A side note: $1 + zeta(2) P(2) = sum_{n=1}^{infty} frac{omega(n)}{n^2}$, where $omega(n)$ is the number of distinct prime divisors of $n$ and $P$ is the prime zeta function. Although you are interested in $sum_{n = 2|omega(n)}^{infty}frac{1}{n^2}$ I wonder if you can utilise some kind of mobius inversion or if you would just go in circles?
$endgroup$
– DanielOnMSE
Dec 6 '18 at 14:59
1
1
$begingroup$
A side note: $1 + zeta(2) P(2) = sum_{n=1}^{infty} frac{omega(n)}{n^2}$, where $omega(n)$ is the number of distinct prime divisors of $n$ and $P$ is the prime zeta function. Although you are interested in $sum_{n = 2|omega(n)}^{infty}frac{1}{n^2}$ I wonder if you can utilise some kind of mobius inversion or if you would just go in circles?
$endgroup$
– DanielOnMSE
Dec 6 '18 at 14:59
$begingroup$
A side note: $1 + zeta(2) P(2) = sum_{n=1}^{infty} frac{omega(n)}{n^2}$, where $omega(n)$ is the number of distinct prime divisors of $n$ and $P$ is the prime zeta function. Although you are interested in $sum_{n = 2|omega(n)}^{infty}frac{1}{n^2}$ I wonder if you can utilise some kind of mobius inversion or if you would just go in circles?
$endgroup$
– DanielOnMSE
Dec 6 '18 at 14:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider the product over all prime numbers
$$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pleft(1-frac{1}{p^2}+frac{1}{(p^2)^2}+dotsright).$$
If you think about how a term of this series will look like when you multiply it out, you will find that there is a term $1/n^2$ for every $n$ with an even number of prime factors, and a term $-1/n^2$ for every $n$ with an odd number of prime factors. Therefore, we find that $P=2S_2-zeta(2)$. Since we know the value of $zeta(2)$ and want to find the value of $S_2$, it's enough to find the value of $P$.
But observe
$$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pfrac{(1-1/p^4)^{-1}}{(1-1/p^2)^{-1}}=frac{prod_p(1-1/p^4)^{-1}}{prod_p(1-1/p^2)^{-1}}=frac{zeta(4)}{zeta(2)}=frac{pi^4/90}{pi^2/6}=frac{pi^2}{15}.$$
Comparing this with the above,
$$S_2=frac{P+zeta(2)}{2}=frac{pi^2/15+pi^2/6}{2}=frac{7pi^2}{60}.$$
answered Dec 6 '18 at 14:46
WojowuWojowu
18.8k23172
$endgroup$
2
$begingroup$
Very nice indeed :)
$endgroup$
– gandalf61
Dec 6 '18 at 14:50
add a comment |
$begingroup$
About the alternative version,
$$begin{eqnarray*} sum_{substack{ngeq 1\ omega(n)text{ is even}}}!!!!!!frac{1}{n^2}&=&frac{1}{2}left[zeta(2)+sum_{ngeq 1}frac{(-1)^{omega(n)}}{n^2}right]=frac{pi^2}{12}+frac{1}{2}prod_{p}left(1-frac{1}{p^2}-frac{1}{p^4}-frac{1}{p^6}-ldotsright)\&=&frac{pi^2}{12}+frac{1}{2}prod_{p}frac{p^2-2}{p^2-1}=frac{pi^2}{12}left[1+prod_{p}left(1-frac{2}{p^2}right)right]=zeta(2)C_{text{Feller-Tornier}}end{eqnarray*} $$
where the Feller-Tornier constant (close to $frac{2}{3}$) is related to the probability for two consecutive integers of being both squarefree.
answered Dec 6 '18 at 16:09
Jack D'AurizioJack D'Aurizio
291k33284666
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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$begingroup$
Consider the product over all prime numbers
$$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pleft(1-frac{1}{p^2}+frac{1}{(p^2)^2}+dotsright).$$
If you think about how a term of this series will look like when you multiply it out, you will find that there is a term $1/n^2$ for every $n$ with an even number of prime factors, and a term $-1/n^2$ for every $n$ with an odd number of prime factors. Therefore, we find that $P=2S_2-zeta(2)$. Since we know the value of $zeta(2)$ and want to find the value of $S_2$, it's enough to find the value of $P$.
But observe
$$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pfrac{(1-1/p^4)^{-1}}{(1-1/p^2)^{-1}}=frac{prod_p(1-1/p^4)^{-1}}{prod_p(1-1/p^2)^{-1}}=frac{zeta(4)}{zeta(2)}=frac{pi^4/90}{pi^2/6}=frac{pi^2}{15}.$$
Comparing this with the above,
$$S_2=frac{P+zeta(2)}{2}=frac{pi^2/15+pi^2/6}{2}=frac{7pi^2}{60}.$$
answered Dec 6 '18 at 14:46
WojowuWojowu
18.8k23172
$endgroup$
2
$begingroup$
Very nice indeed :)
$endgroup$
– gandalf61
Dec 6 '18 at 14:50
add a comment |
$begingroup$
Consider the product over all prime numbers
$$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pleft(1-frac{1}{p^2}+frac{1}{(p^2)^2}+dotsright).$$
If you think about how a term of this series will look like when you multiply it out, you will find that there is a term $1/n^2$ for every $n$ with an even number of prime factors, and a term $-1/n^2$ for every $n$ with an odd number of prime factors. Therefore, we find that $P=2S_2-zeta(2)$. Since we know the value of $zeta(2)$ and want to find the value of $S_2$, it's enough to find the value of $P$.
But observe
$$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pfrac{(1-1/p^4)^{-1}}{(1-1/p^2)^{-1}}=frac{prod_p(1-1/p^4)^{-1}}{prod_p(1-1/p^2)^{-1}}=frac{zeta(4)}{zeta(2)}=frac{pi^4/90}{pi^2/6}=frac{pi^2}{15}.$$
Comparing this with the above,
$$S_2=frac{P+zeta(2)}{2}=frac{pi^2/15+pi^2/6}{2}=frac{7pi^2}{60}.$$
answered Dec 6 '18 at 14:46
WojowuWojowu
18.8k23172
$endgroup$
2
$begingroup$
Very nice indeed :)
$endgroup$
– gandalf61
Dec 6 '18 at 14:50
add a comment |
$begingroup$
Consider the product over all prime numbers
$$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pleft(1-frac{1}{p^2}+frac{1}{(p^2)^2}+dotsright).$$
If you think about how a term of this series will look like when you multiply it out, you will find that there is a term $1/n^2$ for every $n$ with an even number of prime factors, and a term $-1/n^2$ for every $n$ with an odd number of prime factors. Therefore, we find that $P=2S_2-zeta(2)$. Since we know the value of $zeta(2)$ and want to find the value of $S_2$, it's enough to find the value of $P$.
But observe
$$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pfrac{(1-1/p^4)^{-1}}{(1-1/p^2)^{-1}}=frac{prod_p(1-1/p^4)^{-1}}{prod_p(1-1/p^2)^{-1}}=frac{zeta(4)}{zeta(2)}=frac{pi^4/90}{pi^2/6}=frac{pi^2}{15}.$$
Comparing this with the above,
$$S_2=frac{P+zeta(2)}{2}=frac{pi^2/15+pi^2/6}{2}=frac{7pi^2}{60}.$$
answered Dec 6 '18 at 14:46
WojowuWojowu
18.8k23172
$endgroup$
Consider the product over all prime numbers
$$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pleft(1-frac{1}{p^2}+frac{1}{(p^2)^2}+dotsright).$$
If you think about how a term of this series will look like when you multiply it out, you will find that there is a term $1/n^2$ for every $n$ with an even number of prime factors, and a term $-1/n^2$ for every $n$ with an odd number of prime factors. Therefore, we find that $P=2S_2-zeta(2)$. Since we know the value of $zeta(2)$ and want to find the value of $S_2$, it's enough to find the value of $P$.
But observe
$$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pfrac{(1-1/p^4)^{-1}}{(1-1/p^2)^{-1}}=frac{prod_p(1-1/p^4)^{-1}}{prod_p(1-1/p^2)^{-1}}=frac{zeta(4)}{zeta(2)}=frac{pi^4/90}{pi^2/6}=frac{pi^2}{15}.$$
Comparing this with the above,
$$S_2=frac{P+zeta(2)}{2}=frac{pi^2/15+pi^2/6}{2}=frac{7pi^2}{60}.$$
answered Dec 6 '18 at 14:46
WojowuWojowu
18.8k23172
answered Dec 6 '18 at 14:46
WojowuWojowu
18.8k23172
answered Dec 6 '18 at 14:46
WojowuWojowu
18.8k23172
answered Dec 6 '18 at 14:46
WojowuWojowu
18.8k23172
18.8k23172
2
$begingroup$
Very nice indeed :)
$endgroup$
– gandalf61
Dec 6 '18 at 14:50
add a comment |
2
$begingroup$
Very nice indeed :)
$endgroup$
– gandalf61
Dec 6 '18 at 14:50
2
2
$begingroup$
Very nice indeed :)
$endgroup$
– gandalf61
Dec 6 '18 at 14:50
$begingroup$
Very nice indeed :)
$endgroup$
– gandalf61
Dec 6 '18 at 14:50
add a comment |
$begingroup$
About the alternative version,
$$begin{eqnarray*} sum_{substack{ngeq 1\ omega(n)text{ is even}}}!!!!!!frac{1}{n^2}&=&frac{1}{2}left[zeta(2)+sum_{ngeq 1}frac{(-1)^{omega(n)}}{n^2}right]=frac{pi^2}{12}+frac{1}{2}prod_{p}left(1-frac{1}{p^2}-frac{1}{p^4}-frac{1}{p^6}-ldotsright)\&=&frac{pi^2}{12}+frac{1}{2}prod_{p}frac{p^2-2}{p^2-1}=frac{pi^2}{12}left[1+prod_{p}left(1-frac{2}{p^2}right)right]=zeta(2)C_{text{Feller-Tornier}}end{eqnarray*} $$
where the Feller-Tornier constant (close to $frac{2}{3}$) is related to the probability for two consecutive integers of being both squarefree.
answered Dec 6 '18 at 16:09
Jack D'AurizioJack D'Aurizio
291k33284666
$endgroup$
add a comment |
$begingroup$
About the alternative version,
$$begin{eqnarray*} sum_{substack{ngeq 1\ omega(n)text{ is even}}}!!!!!!frac{1}{n^2}&=&frac{1}{2}left[zeta(2)+sum_{ngeq 1}frac{(-1)^{omega(n)}}{n^2}right]=frac{pi^2}{12}+frac{1}{2}prod_{p}left(1-frac{1}{p^2}-frac{1}{p^4}-frac{1}{p^6}-ldotsright)\&=&frac{pi^2}{12}+frac{1}{2}prod_{p}frac{p^2-2}{p^2-1}=frac{pi^2}{12}left[1+prod_{p}left(1-frac{2}{p^2}right)right]=zeta(2)C_{text{Feller-Tornier}}end{eqnarray*} $$
where the Feller-Tornier constant (close to $frac{2}{3}$) is related to the probability for two consecutive integers of being both squarefree.
answered Dec 6 '18 at 16:09
Jack D'AurizioJack D'Aurizio
291k33284666
$endgroup$
add a comment |
$begingroup$
About the alternative version,
$$begin{eqnarray*} sum_{substack{ngeq 1\ omega(n)text{ is even}}}!!!!!!frac{1}{n^2}&=&frac{1}{2}left[zeta(2)+sum_{ngeq 1}frac{(-1)^{omega(n)}}{n^2}right]=frac{pi^2}{12}+frac{1}{2}prod_{p}left(1-frac{1}{p^2}-frac{1}{p^4}-frac{1}{p^6}-ldotsright)\&=&frac{pi^2}{12}+frac{1}{2}prod_{p}frac{p^2-2}{p^2-1}=frac{pi^2}{12}left[1+prod_{p}left(1-frac{2}{p^2}right)right]=zeta(2)C_{text{Feller-Tornier}}end{eqnarray*} $$
where the Feller-Tornier constant (close to $frac{2}{3}$) is related to the probability for two consecutive integers of being both squarefree.
answered Dec 6 '18 at 16:09
Jack D'AurizioJack D'Aurizio
291k33284666
$endgroup$
About the alternative version,
$$begin{eqnarray*} sum_{substack{ngeq 1\ omega(n)text{ is even}}}!!!!!!frac{1}{n^2}&=&frac{1}{2}left[zeta(2)+sum_{ngeq 1}frac{(-1)^{omega(n)}}{n^2}right]=frac{pi^2}{12}+frac{1}{2}prod_{p}left(1-frac{1}{p^2}-frac{1}{p^4}-frac{1}{p^6}-ldotsright)\&=&frac{pi^2}{12}+frac{1}{2}prod_{p}frac{p^2-2}{p^2-1}=frac{pi^2}{12}left[1+prod_{p}left(1-frac{2}{p^2}right)right]=zeta(2)C_{text{Feller-Tornier}}end{eqnarray*} $$
where the Feller-Tornier constant (close to $frac{2}{3}$) is related to the probability for two consecutive integers of being both squarefree.
answered Dec 6 '18 at 16:09
Jack D'AurizioJack D'Aurizio
291k33284666
answered Dec 6 '18 at 16:09
Jack D'AurizioJack D'Aurizio
291k33284666
answered Dec 6 '18 at 16:09
Jack D'AurizioJack D'Aurizio
291k33284666
answered Dec 6 '18 at 16:09
Jack D'AurizioJack D'Aurizio
291k33284666
291k33284666
add a comment |
add a comment |
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$begingroup$
A side note: $1 + zeta(2) P(2) = sum_{n=1}^{infty} frac{omega(n)}{n^2}$, where $omega(n)$ is the number of distinct prime divisors of $n$ and $P$ is the prime zeta function. Although you are interested in $sum_{n = 2|omega(n)}^{infty}frac{1}{n^2}$ I wonder if you can utilise some kind of mobius inversion or if you would just go in circles?
$endgroup$
– DanielOnMSE
Dec 6 '18 at 14:59