# Convergent sum involving the divisor function

Here's a doosy posed as a challenge question;

Let $d(n)$ be the number of divisors of $n$. Prove that $d(n)$ is a multiplicative function of $n$ and show that for any natural number $r \ge 1$, the series

$$\sum_{n=1}^{\infty}{\frac{d(n)^r}{n^s}}$$ converges absolutely for $s \gt 1$. Deduce that for any $\epsilon \gt 0, d(n) = O(n^{\epsilon}).$

I'm completely stuck. When I see $O$ terms, I automatically think partial summation, but I don't see that working here. Not sure how to approach this, any help is appreciated

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Hint: 1. Multiplicativity of $d(n)$ and $d(n)^r$ should be do-able. Try to do that first. 2. Factor the series into Euler products, as you would for the Riemann zeta function. Can you see convergence? 3. Once you know the series converges for $s > 1$, it means that each term has to converge to zero eventually, no matter how large $r$ is. Can you deduce anything about the size of $d(n)$? –  user27126 Nov 5 '12 at 19:06

Your series is related to the Dirichlet series.

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this was a great point in the right direction. thanks! –  user45814 Nov 5 '12 at 19:08
@user45814: You are welcome. –  Mhenni Benghorbal Nov 12 '12 at 4:50