# Riemann Zeta Identity

• Jun 7th 2010, 05:37 PM
chiph588@
Riemann Zeta Identity
For $\displaystyle \Re(s)>1$ show $\displaystyle \sum_{n=1}^\infty \frac{\tau\left(n^2\right)}{n^s} = \frac{\zeta^3(s)}{\zeta(2s)}$ where $\displaystyle \tau(m)$ is the number of divisors of $\displaystyle m$.
• Jun 7th 2010, 08:39 PM
simplependulum
If $\displaystyle m = p_1^{a_1}p_2^{a_2}p_2^{a_3}...$ then $\displaystyle \tau(m) = (1 + a_1)(1 + a_2 )( 1 + a_3) ...$

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

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

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

Since

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

By taking derivative ,

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

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

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

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

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

Therefore ,

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

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

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