# 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.

• is that second summation really in the exponent? – T_M Nov 16 '18 at 18:38
• 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. – SmileyCraft Nov 16 '18 at 18:44
• 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 $\infty \cdot -\infty$. – Reese Nov 16 '18 at 18:46
• Sorry, it is $1/n^2$, I forgot the ^2 – Arsh Nov 16 '18 at 18:48
• Please don't use display style in title. I've edited this away. – 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 $$n\to\infty$$.

• Why does your answer differ by a minus sign, have I done anything wrong? – Zacky Nov 16 '18 at 19:36
• @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. – 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 $$k=\pi^2/6$$.