we remember that \[\sum_{n=1}^\infty\dfrac{Ch(n)}{n^s}=-2\zeta(2s)+\zeta(s)\] consider \[Cs(n)=(Ch*1)(n)=\sum_{d|n}Ch(d)\] so, if d is a perfect square it is -1, 1 otherwise. if there are k perfect square divisors, there are \(\tau(n)-k\) non-square divisor. add them and it is \(\tau(n)-2k\).

let k be a function of n. It is multiplicative. So we can compute for **prime powers**
\[k(p^a)=\sum_{d^2|p^a} d^{0}=1^0+(p^2)^0+(p^4)^0+.... \left(p^{2\lfloor \dfrac{a}{2}\rfloor}\right)^0=\lfloor \dfrac{a}{2}\rfloor+1\]
we have
\[\sum_{n=1}^\infty\dfrac{Ch(n)}{n^s}\sum_{n=1}^\infty\dfrac{1}{n^s}=\sum_{n=1}^\infty\dfrac{\tau(n)}{n^s}-2\sum_{n=1}^\infty\dfrac{k(n)}{n^s}\]
\[-2\zeta(s)\zeta(2s)+\zeta^2(s)=\zeta^2(s)-2\sum_{n=1}^\infty\dfrac{k(n)}{n^s}\]
\[\sum_{n=1}^\infty\dfrac{k(n)}{n^s}=\zeta(s)\zeta(2s)\]
Lets try to compute the summation using the fact it is multpicative
\[\prod_{p=prime} \sum_{a≥0}\dfrac{k(p^a)}{p^{as}}=\prod_{p=prime} \sum_{a≥0}\dfrac{\lfloor \dfrac{a}{2}\rfloor+1}{p^{as}}=\prod_{p=prime} \dfrac{p^{3s}}{(p^s-1)^2(p^s+1)}\\=\prod_p \dfrac{1}{(1-p^{-s})^2}\dfrac{1-p^ {-s}}{1-p^{-2s}}=\zeta^2(s)\dfrac{\zeta(2s)}{\zeta(s)}=\zeta(s)\zeta(2s)\]
so we get
\[\boxed{\large{s=s}}\]
again!!!

## Comments

Sort by:

TopNewestThe combo part was not found by me but @Andrew Ellinor so he gets the credit for that.

He was not sure wether or not he was correct but this proves that \[k(n)=\prod_{i=1}^{\omega(n)} (\lfloor \dfrac{k_i}{2}\rfloor +1)\] if \(n=p_1^{k_1}p_2^{k_2}.....p_{\omega(n)}^{k_{\omega(n)}}\) – Aareyan Manzoor · 1 year, 7 months ago

Log in to reply

– Julian Poon · 1 year, 7 months ago

Well he is correct.Log in to reply