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Im trying to show, for $\Re(s)>1$, that $\displaystyle\sum_{n=0}^{\infty} \frac{d(n^2)}{n^s} = \frac{\zeta^3(s)}{\zeta(2s)}$, where $d(n)= |\{k \mid k|n \}|$, number of positive integers that divides $n$.

I tried to separate the RHS to $\displaystyle{\zeta^2(s)}=\sum_{n=0}^{\infty} \frac{u* u(n)}{n^s}=\sum_{n=0}^{\infty} \frac{d(n)}{n^s}$ which obtained by the dirichlet sum of $\zeta (s)$ and $\dfrac{\zeta(s)}{\zeta(2s)}$ which I tried to simplify by Euler Product formula but it didnt came up to something.

Could you please help me find a way of handling that?

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The Euler product of $\zeta(s)^3/\zeta(2s)$ is $$\frac{\zeta(s)^3}{\zeta(2s)} = \prod_p \frac{1-p^{-2s}}{(1-p^{-s})^3} = \prod_p \frac{1+p^{-s}}{(1-p^{-s})^2}.$$ But $$\frac{1+p^{-s}}{(1-p^{-s})^2} = \sum_{k=0}^n (2k+1)p^{-ks} = \sum_{k=0}^n d(p^{2k})p^{-ks}.$$ Since $d$ is multiplicative, you should be able to conclude.

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  • $\begingroup$ Could you please explain the transition from $$\frac{1+p^{-s}}{(1-p^{-s})^2}$$ to the infinite sum? $\endgroup$ – Snufsan Jul 30 '14 at 12:58
  • $\begingroup$ @Snufsan: Note that the denominator is the square of an infinite series, and so has a tidy expression as a sum. Multiplying by the numerator then adds to that series a 'shifted' copy of itself, and that gives the simple form shown. $\endgroup$ – Semiclassical Jul 30 '14 at 13:15
  • $\begingroup$ Still i cant figure how we got the $2k+1$ factor. we have $$\frac{1}{(1-p^{-s})^2} = \sum_{n=0}^{\infty} \frac{d(p^k)}{p^{ks}} $$ and then what? $\endgroup$ – Snufsan Jul 30 '14 at 13:21
  • $\begingroup$ Nevermind, Got it! $\endgroup$ – Snufsan Jul 30 '14 at 13:34

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