Why does $\sum_{p\in P} \sum_{k\in N^+} \frac {1}{k.^{2}} $ yield $\ln(\frac{\pi^2}{6})$?

Question

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!

Answer 1

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 $s=2$.