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The exact series I must show converges absolutely is:

$$\sum_{n=1}^{\infty}{\frac{d(n)^r}{n^s}}$$

for $s > 1$, $r\in \mathbb{N}$ and where $d(n)=\#\text{ of divisors of } n$. I've been able to show that $d(n)$, $d(n)^r$ are multiplicative and so my series is Dirichlet. As such, I've broken this into Euler sums and thus transformed the series to

$$\prod_{P}\left(1 + \frac{d(p)^r}{p^s} + \frac{d(p^2)^r}{p^{2s}} + \cdots\right) = \prod_{P}\sum_{s=0}^{\infty}{\frac{(s+1)^r}{p^s}}$$

I'm not sure how to proceed from here. Any help is appreciated, thanks!

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  • $\begingroup$ First, your infinite series in Euler product is wrong. It should be $\sum_n \frac{(n+1)^r}{p^{ns}}$. $\endgroup$ – user27126 Nov 14 '12 at 1:07
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    $\begingroup$ One way to proceed is to take log and bound the log of each term in the Euler product. Try to show that each series $\sum_n \frac{(n+1)^r}{p^{ns}}$ is really a sum of finitely many terms (depending on $r$) of constants times $\frac{1}{(log p)^k(1-p^{-s})^k}$, where $k$ goes up to $r$. This allows you to bound log of the series, by sum of log of $\frac{1}{(log p)^k(1-p^{-s})^k}$, and it's easy to see that the latter sum converges for $s > 1$. $\endgroup$ – user27126 Nov 14 '12 at 1:09
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It is known (see for instance Apostol's book Introduction to analytic number theory) that for all $\epsilon>0$ $$ d(n)=o(n^\epsilon). $$ Given $s>1$ choose $\epsilon=\dfrac{s-1}{2\,r}>0$. Then $$ \sum_{n=1}^\infty\frac{d(n)^r}{n^s}\le C\sum_{n=1}^\infty\frac{n^{r\epsilon}}{n^s}=C\sum_{n=1}^\infty\frac{1}{n^{s-r\epsilon}}=C\sum_{n=1}^\infty\frac{1}{n^{(s+1)/2}}<\infty, $$ where $C$ is a constant depending on $s$ and $r$.

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  • $\begingroup$ I was going to include it in the question, but decided against it because it would just add clutter. But the relation $d(n)=O(n^ϵ)$ is what we ultimately need to deduce $\endgroup$ – user45793 Nov 12 '12 at 18:55

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