How to evaluate the limit $\lim_{m\to\infty}m\left({(\sum_{n=1}^{m}\frac{1}{n^2})}^{\pi^2/6}-{(\pi^2/6)}^{\sum_{n=1}^{m}\frac{1}{n^2}}\right)$ 
How to evaluate the following limit? 
  $$
\lim_{m\to\infty}m\left[
\left(\sum_{n=1}^{m}\frac{1}{n^2}\right)^{\pi^2/6}-{(\pi^2/6)}^{\sum_{n=1}^{m}\frac{1}{n^2}}
\right]
$$

This limit is of the form $\infty \cdot 0$. I generally solve such problem by taking the term that equals $0$ to the denominator and then using l'Hôpital rule. However, that won't work here.
 A: Write $c = \frac{\pi^2}{6} = \sum_{k=1}^{\infty} \frac{1}{k^2}$ and $\epsilon_n = \sum_{k=n+1}^{\infty} \frac{1}{k^2}$. Then


*

*$ c^{c - \epsilon_n} = c^c \left[ 1 - \epsilon_n \log c + \mathcal{O}(\epsilon_n) \right] $,

*$ (c - \epsilon_n)^c = c^c \left( 1 - \frac{\epsilon_n}{c} \right)^c = c^c \left[ 1 - \epsilon_n + \mathcal{O}(\epsilon_n^2) \right] $,

*$\epsilon_n = \int_{n}^{\infty} \frac{dx}{x^2} + \mathcal{O}\left(\frac{1}{n^2}\right) = \frac{1}{n} + \mathcal{O}\left(\frac{1}{n^2}\right) $.
Combining altogether,
$$ n \left( c^{c-\epsilon_n} - (c-\epsilon_n)^c \right) = c^c (1 - \log c) + \mathcal{O}\left(\frac{1}{n}\right), $$
which converges to $c^c (1 - \log c) = \zeta(2)^{\zeta(2)} \left( 1 - \log \zeta(2) \right)$ as $n\to\infty$.
A: The factor $m$ can be moved into denominator as $1/m$ and then we can apply Cesaro-Stolz. The numerator of the expression after applying Cesaro-Stolz is $$(a+h) ^k-a^k-k^{a+h} +k^a$$ where $$k=\frac{\pi^2}{6},a=\sum_{i=1}^{m}i^{-2}\to k,h=(m+1)^{-2}\to 0$$ Thus $k$ is constant and $a, h$ are functions of $m$. The denominator of the expression after applying Cesaro-Stolz is $-1/m(m+1)$ and this can be replaced by $-h$. Thus we need to find the limit $$\lim_{h\to 0}\frac{a^k-(a+h)^k+k^{a+h}-k^a}{h}$$ and this is easily evaluated as $k^k\log k-k^k$ where $k=\pi^2/6$.
