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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.

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$\sigma_k(n)=\sum_{d\mid n}d^k$ 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.

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  • $\begingroup$ For which $s$ this holds? $\endgroup$
    – user365151
    Commented Sep 6, 2016 at 18:45
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    $\begingroup$ @user365151: as soon as the involved series are absolutely convergent, hence for $\text{Re}(s)>k+1$. $\endgroup$
    – Jack D'Aurizio
    Commented Sep 6, 2016 at 18:48
  • $\begingroup$ 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? $\endgroup$
    – user365151
    Commented Sep 6, 2016 at 19:24
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    $\begingroup$ @user365151: exactly. $\endgroup$
    – Jack D'Aurizio
    Commented Sep 6, 2016 at 20:12
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    $\begingroup$ @user365151: there is no big difference, since if $s=\sigma+it$ we have $$\left\|\frac{1}{n^s}\right\|=\frac{1}{\|n^\sigma\cdot n^{it}\|}=\frac{1}{n^\sigma}.$$ $\endgroup$
    – Jack D'Aurizio
    Commented Sep 6, 2016 at 21:05

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