# Dirichlet series of the divisor function

I want to show that the Dirichlet series of the divisor function $\sigma_k$ converges absolutly.

I have found this: $$\sum_{n=1}^{\infty}\frac{\sigma_{k}(n)}{n^{s}}=\zeta(s)\zeta(s-k)$$ for all $s\in\mathbb{C}$ with $\text{Re}(s)>k+1$.

I thought: The series convergs absolutly as a product of absolutly convergent series (zeta-series). Is that right?

If yes: Where can I find a proof of this equation? If no: How can this be shown?

Thanks.

## 1 Answer

$\sigma_k(n)=\sum_{d\mid n}d^k$ is a multiplicative function, hence by Euler's product or just by convolution of Dirichlet's series we have

$$\sum_{n\geq 1}\frac{\sigma_k(n)}{n^s}=\sum_{n\geq 1}\frac{1}{n^s}\sum_{n\geq 1}\frac{n^k}{n^s}=\zeta(s)\zeta(s-k)$$

as soon as the involved series are absolutely convergent.

• For which $s$ this holds? – user365151 Sep 6 '16 at 18:45
• @user365151: as soon as the involved series are absolutely convergent, hence for $\text{Re}(s)>k+1$. – Jack D'Aurizio Sep 6 '16 at 18:48
• Thank you! So the left series is absolutely convergent because the right side is absolutely convergent because it is a product of absolutely convergent zeta series? – user365151 Sep 6 '16 at 19:24
• @user365151: exactly. – Jack D'Aurizio Sep 6 '16 at 20:12
• @user365151: there is no big difference, since if $s=\sigma+it$ we have $$\left\|\frac{1}{n^s}\right\|=\frac{1}{\|n^\sigma\cdot n^{it}\|}=\frac{1}{n^\sigma}.$$ – Jack D'Aurizio Sep 6 '16 at 21:05