Riemann Zeta Identity

**chiph588@** · June 7th 2010, 05:37 PM

For $\Re(s)>1$ show $\sum_{n=1}^\infty \frac{\tau\left(n^2\right)}{n^s} = \frac{\zeta^3(s)}{\zeta(2s)}$ where $\tau(m)$ is the number of divisors of $m$ .

**simplependulum** · June 7th 2010, 08:39 PM

If $m = p_1^{a_1}p_2^{a_2}p_2^{a_3}...$ then $\tau(m) = (1 + a_1)(1 + a_2 )( 1 + a_3) ...$

and $\tau(m^2) = (1 + 2a_1 )( 1 + 2a_2 )( 1 + 2a_3) ...$

and the sum $\sum_{n=1}^{\infty} \frac{\tau(n^2)}{n^s}$

$= \prod_{p_i \in \mathbb{P} }\left( \sum_{k=0}^{\infty} \frac{2k+1}{p_i^{ks}} \right)$

Since

$\frac{1}{1-x} = 1 + x + x^2 + ....$

By taking derivative ,

$\frac{x}{(1-x)^2} = x + 2x^2 + 3x^3 + ...$

Then $\sum_{k=0}^{\infty} (2k+1)x^k = \frac{2x}{(1-x)^2} + \frac{1}{1-x} = \frac{2x+1-x}{(1-x)^2} = \frac{1 + x}{(1-x)^2}$

Let $x = \frac{1}{p_i^s }$

we have $\sum_{k=0}^{\infty} \frac{2k+1}{p_i^{ks}} = \frac{1 + 1/p_i^s }{ \left( 1 - \frac{1}{p_i^s} \right)^2 }$

$= \frac{1 - 1/p_i^{2s} }{ \left( 1 - \frac{1}{p_i^s} \right)^3 }$

Therefore ,

$\sum_{n=1}^{\infty} \frac{\tau(n^2)}{n^s} = \prod_{p_i \in \mathbb{P} } \frac{1 - 1/p_i^{2s} }{ \left( 1 - \frac{1}{p_i^s} \right)^3 }$ $= \frac{ \zeta^3(s)}{\zeta(2s)}$ since

$\frac{1}{ \zeta(z)} = \prod_{p_i \in \mathbb{P} } (1- \frac{1}{p_i^z} )$

* $\mathbb{P}$ denotes the set of all prime numbers .

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