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.

  • $\begingroup$ is that second summation really in the exponent? $\endgroup$ – T_M Nov 16 '18 at 18:38
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    $\begingroup$ It probably helps to know that $\sum_{n=1}^m\frac1n-\ln(m)\to\gamma$ where $\gamma$ is a constant. The Euler Mascheroni constant to be precise. $\endgroup$ – SmileyCraft Nov 16 '18 at 18:44
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    $\begingroup$ Do you mean $1/n^2$ instead of $1/n$? Otherwise the latter part of that difference grows much faster than the former, and you have a limit of the form $\infty \cdot -\infty$. $\endgroup$ – Reese Nov 16 '18 at 18:46
  • $\begingroup$ Sorry, it is $1/n^2$, I forgot the ^2 $\endgroup$ – Arsh Nov 16 '18 at 18:48
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    $\begingroup$ Please don't use display style in title. I've edited this away. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Nov 16 '18 at 19:01

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 $n\to\infty$.

  • $\begingroup$ Why does your answer differ by a minus sign, have I done anything wrong? $\endgroup$ – Zacky Nov 16 '18 at 19:36
  • $\begingroup$ @Zacky, I made a mistake by switching two terms at the beginning. My answer itself is not wrong, but it computed the negative of OP's question. $\endgroup$ – Sangchul Lee Nov 16 '18 at 19:40

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 $k=\pi^2/6$.


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