# Hint on a limit that involves the Hurwitz Zeta function

I will be honest. Some play with a weird integral has gotten me to this formulation: $$\lim\limits_{n\mathop\to\infty}\frac{\zeta(2,n)}{\frac 1n+\frac 1{2n^2}}=1$$ It seems true because of the numerical approximations made and the actual evaluation of this limit at Wolfram Alpha. But i am stuck. Does anyone have an idea of where to start?

-
Do you consider accepting answers? – Mhenni Benghorbal Aug 31 '14 at 1:02
Of course i will! – Daniel Charry Aug 31 '14 at 1:03
The idea is correct. I made a correction. The limit should be one. – Mhenni Benghorbal Aug 31 '14 at 2:56

We have that: $$\zeta(2,n)=\int_{0}^{+\infty}\frac{t\,e^{-nt}}{1-e^{-t}}\,dt$$ but since the integral on $[\pi,+\infty)$ is very small (exponentially small in $n$) we can just consider the Taylor series of $\frac{t}{1-e^{-t}}$ with respect to the point $t=0$ in order to have: $$\begin{eqnarray*}\zeta(2,n)&=&\int_{0}^{\pi}\left(1+\frac{t}{2}+\frac{t^2}{12}+O(t^4)\right)e^{-nt}\,dt + O(e^{-n})\\&=&\int_{0}^{+\infty}\left(1+\frac{t}{2}+\frac{t^2}{12}\right)e^{-nt}\,dt+O\left(\frac{1}{n^5}\right)\\&=&\frac{1}{n}+\frac{1}{2n^2}+\frac{1}{6n^3}+O\left(\frac{1}{n^5}\right),\end{eqnarray*}$$ that is much stronger than the limit it is asked to prove.

-
+1 Cool, now I can upvote this. – achille hui Aug 31 '14 at 3:05

If $\zeta(2,n)$ gives the generalized Riemann zeta function, for large values of $n$, the Taylor expansion is $$\zeta(2,n)=\frac{1}{n}+\frac{1}{2 n^2}+\frac{1}{6 n^3}-\frac{1}{30 n^5}+O\left(\left(\frac{1}{n}\right)^7\right)$$ So,$$\frac{\zeta(2,n)}{\frac 1n+\frac 1{2n^2}}=1+\frac{1}{6 n^2}-\frac{1}{12 n^3}+\frac{1}{120 n^4}-\frac{1}{240 n^5}+\frac{29}{1120 n^6}+O\left(\left(\frac{1}{n}\right)^7\right)$$ and then $$\lim\limits_{n\mathop\to\infty}\frac{\zeta(2,n)}{\frac 1n+\frac 1{2n^2}}=1$$

-
Just to be a major nitpicker: how do you prove that $\zeta(2,n)$ has such Taylor series? – Jack D'Aurizio Aug 31 '14 at 3:07
@JackD'Aurizio apply Watson's Lemma to the integral representation of $\zeta(2,n)$ appeared in your answer? – achille hui Aug 31 '14 at 3:13
@achillehui: yes, for sure ;) I was just debating answers like "in virtue of the following unproven relation, your limit equals..." – Jack D'Aurizio Aug 31 '14 at 3:17

You can use the asymptotic expansion of $\zeta(2,n)$

$$\frac{1}{n}+ \frac{2}{2n^2}+O(\frac{1}{n^3}) .$$

Added: I provided my answer based on your request for a hint. However here is one technique you can use.

-
Just to be a major nitpicker: how do you prove this identity? – Jack D'Aurizio Aug 31 '14 at 3:06
@JackD'Aurizio: The OP asked for a hint! However you can find some techniques that I used in different answers if you are interested. So the main idea for solving the problem was given in this answer. – Mhenni Benghorbal Aug 31 '14 at 3:09
you just posted a shortened version of C.Leibovici's answer, what is the point in writing it? The OP already had such a hint in a previously posted answer. – Jack D'Aurizio Aug 31 '14 at 3:11