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?
 A: 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.
A: 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 $$\lim\limits_{n\mathop\to\infty}\frac{\zeta(2,n)}{\frac 1n+\frac 1{2n^2}}=1$$
A: You can use the asymptotic expansion of $\zeta(2,n)$

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

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