Riemann Zeta Identity

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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 \(\displaystyle m \).
 

simplependulum

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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 .
 
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