I have been looking at this for days and I am going insane. 


I need to show that for a Dirichlet series equal to $\zeta(s)\zeta(2s)$, the sum of the coefficients less than $x$ is $x\zeta(2)+O(x^{(3/4)})$, and then expand that to the $\Pi \zeta(ks)$ for all $k$ in an effort to find the formula for the number of non-isomorphic abelian groups.


I know that using Perron's formula there is a simple pole at $s=1$ that gives a residue of $X\zeta(2)$, but I can't find a contour that converges or the exact error term.



Contour integration of $\zeta(s)\zeta(2s)$
This case, at least, you can do by hand; if $c(n)$ is the $n$th coefficient of $\zeta(s)\zeta(2s)$, then


$\sum\_{n\leq X}c(n)=\sum\_{nm^2\leq X}1=\sum\_{m}\sum\_{n\leq X/m^2}1 = \sum\_{m < \sqrt{X}}(m^{-2}
X+O(1))=\zeta(2)X+O(\sqrt{X}).$


