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!
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 $r$.