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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.

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  • $\begingroup$ Let $C(x)$ be the summatory function. One way to handle convergence issues, consider the integral $I(x)=\int_1^x C(t) dt$. Then $I(x)=\int \zeta(s)\zeta(2s){x^{s+1} ds\over s(s+1)}$. Move to $Re(s)=\epsilon$, integral converges ($t$-aspect growth of $\zeta$), get poles at $s=1,1/2$, so $I(x)=cx^2+dx^{3/2}+O_\epsilon(x^{1+\epsilon})$. Then go back to $C(x)$, essentially via differentation with care in error term. For general case you can follow same track, with iterated integrals as one idea to demand convergence. $\endgroup$
    – Junkie
    Commented Mar 18, 2011 at 5:07

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

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

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  • $\begingroup$ Thank you. If anyone knows how to extend this to $\Pi\zeta(ks)$ it would be very much appreciated $\endgroup$
    – AbelianGrapes
    Commented Mar 16, 2011 at 5:09
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    $\begingroup$ @unknown: The sum of coefficients of $\prod_{k=1}^K\zeta(ks)$ up to $X$ is $x\prod_{k=2}^K\zeta(k)+O_K(\sqrt{x})$. This can be seen by induction on $K$, using a similar calculation as the response above. $\endgroup$
    – GH from MO
    Commented Mar 16, 2011 at 15:43
  • $\begingroup$ I'm still trying to solve this using perron's formula. I am using cauchy's theorm for c=3/4 but I'm having some trouble bounding this integral. $\endgroup$
    – AbelianGrapes
    Commented Mar 17, 2011 at 2:28
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    $\begingroup$ Try using the approximate Perron formula $h(x)=\frac{1}{2\pi i} \int_{\sigma -iT}^{\sigma + iT}x^s \frac{ds}{s}+O(T^{-1}x^{\sigma})$ and the convexity bound for the zeta function. $\endgroup$
    – David Hansen
    Commented Mar 17, 2011 at 3:57
  • $\begingroup$ Sorry, $h(x)$ is $1$ if $x<1$, $1/2$ if $x=1$, and $0$ if $x>1$. $\endgroup$
    – David Hansen
    Commented Mar 17, 2011 at 3:58
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If you havn't done so already, you might find it useful to look at the proof of theorem 12.2 on the divisor problem in Titchmarsh - The theory of the Riemann zeta function. Here, he goes through a detailed application of Perron's formula for the function $\zeta^k(s)$, which I believe to be very similar to your case.

Indeed, for $s=\sigma + it$ and $\sigma>1/2$, $\zeta(2s)$ is absolutley convergent and hence uniformly bounded with respect to $t$. So this will not contribute to the contours that you choose (as long as $\sigma>1/2$!). What you need then is good upper bounds for the order of the zeta function in the critical strip.

To get these, one normally finds the order of the function at two points, and then uses the Phragmén–Lindelöf principle for strips to get estimates for the function between these two points. For example, it is known that $\zeta(1/2 + it) = O(t^{1/4})$ (see The Lindelöf hypothesis), although there are much better bounds available than that. This is all done in Titchmarsh's book.

I hope this helps!

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