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The source for this problem is this 3b1b video. I understand this:

$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...=\frac{\pi^2}{6}$$

Now he alters the series to include only primes and powers of primes (eg. 4 and 8 are included because they are powers of 2, which is prime) while scaling down the powers of primes by a factor of the exponent, as in:

$$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2(2)}+\frac{1}{5^2}+\frac{1}{7^2}+\frac{1}{8^2(3)}+\frac{1}{9^2(2)}+...$$

This happens to equate to: $$ \ln\left(\frac{\pi^2}{6}\right)$$

I tried to search the proof of this for a while, but could not find anything. I would be delighted to see an elementary explanation to this,

Thanks!

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    $\begingroup$ Is the summation meant to be $$\sum_{p\in\mathcal{P}}\sum_{k\in\mathbb{N}^+}\frac{1}{kp^{2k}}$$? $\endgroup$ Sep 29, 2020 at 5:48
  • $\begingroup$ I don't understand what the series is. Why is $\frac{1}{8^2(3)}$ included, for example? $\endgroup$
    – pancini
    Sep 29, 2020 at 5:49
  • $\begingroup$ @Mark yes, that is a better way to put it, thanks. $\endgroup$
    – Bipasha
    Sep 29, 2020 at 5:51
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    $\begingroup$ @Elliot G Yes, really sorry. That was a typing mistake. $\endgroup$
    – Bipasha
    Sep 29, 2020 at 5:54
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    $\begingroup$ @DavidKipper I think the connection is that for $n = p^k$, $\Lambda(n) / \log(n) = \log(p) / k\log(p) = 1/k$, so it seems to be precisely the same sum. $\endgroup$ Sep 29, 2020 at 6:00

1 Answer 1

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The Euler product factorization of the Riemann zeta function is:

$$ \zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_p \left(1-\frac{1}{p^{s}}\right)^{-1} $$

Apply natural logarithm with the Newton-Mercator expansion

$$ \ln\zeta(s)=\sum_p -\ln\left(1-\frac{1}{p^s}\right)=\sum_p \sum_{k=1}^{\infty}\frac{1}{kp^{ks}}. $$

This is the special case $s=2$.

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    $\begingroup$ In the first line, you have a product over $p$, but no $p$ appears in the formula. Looks like it should be $\prod_p(1-\frac{1}{p^s})^{-1}$? $\endgroup$
    – Alex Ortiz
    Sep 30, 2020 at 1:50

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