mathimatikoi.org

Let $\sigma(n)$ denote the sum of of all divisors of $n$, that is $\displaystyle {\rm \sigma(n)=\sum \limits_{d \mid n} d}$. Prove that for $s \in \mathbb{R} \mid s>2$ it holds that:

$$\sum_{n=1}^{\infty} \frac{\sigma(n)}{n^s} = \zeta(s) \zeta(s-1)$$

where $\zeta$ is the Riemann zeta function.

This is not a casual number thoery problem rather an analytic number theory problem. I found it very interesting and thus I am sharing it with you.

Let $\displaystyle F(z)=\sum_{m=1}^{\infty} \frac{f(m)}{m^z}$ and $\displaystyle G(z)=\sum_{n=1}^{\infty} \frac{g(n)}{n^z}$ be two complex series that converge absolutely somewhere in the complex plane then we define the convolution Dirichlet product as follows:

$$F(z)G(z) = \sum_{m=1}^{\infty} \frac{f(m)}{m^z} \sum_{n=1}^{\infty} \frac{g(n)}{n^z} = \sum_{n=1}^{\infty} \frac{h(n)}{n^z} = \sum_{n=1}^{\infty} \frac{\left ( f*g \right )(n)}{n^z} = \sum_{n=1}^{\infty} \frac{\sum \limits_{{\rm d\mid n}} f(d) g\left ( \frac{n}{d} \right )}{n^z}$$

So taking $f(n)=1, \; g(n)=n$ we have that the convolution product is actually $\sigma(n)$ thus:

$$\sum_{m=1}^{\infty} \frac{1}{m^s} \sum_{n=1}^{\infty} \frac{n}{n^{s}} = \sum_{n=1}^{\infty}\frac{\sigma(n)}{n^s} \Rightarrow \sum_{n=1}^{\infty}\frac{\sigma(n)}{n^s} = \zeta(s) \zeta(s-1)$$

that is what we wanted.

Remark: Extending the result we may obtain that for $s>a$ and $\displaystyle {\rm \sigma_a(n)=\sum \limits_{d \mid n^a} d}$ the following result holds:

$$\sum_{n=1}^{\infty} \frac{\sigma_a(n)}{n^s} = \zeta(s) \zeta(s-a)$$