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How to calculate the limit $$\lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{1}{n^2}-\frac{\pi^2}{6}\right)N?$$ By using the numerical method with Python, I guess the right answer is $-1$ but how to prove? I have no idea. enter image description here

Stephen
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    Do you know about the Basel Problem? – Shimura Variety Apr 28 '22 at 00:35
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    You might be able to use the Euler–Maclaurin summation formula. Wikipedia gives one of the forms related to $\zeta(2)$ as $\sum_{n=1}^N \frac{1}{n^2} \sim \frac{\pi^2}{6}-\frac{1}{n}+\frac{1}{2n^2}-\frac{1}{6n^3}+\frac{1}{30n^5}-\frac{1}{42n^7}+ \cdots $ The problem is you get an asymptotic series that doesn't actually converge. With a little care estimating the remainder, that could be a viable approach. – sharding4 Apr 28 '22 at 00:53
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    A version of Cesaro-Stolz helps to evaluate the limit easily. – Paramanand Singh Apr 28 '22 at 01:05
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    @sharding4 I think that's the right approach. I don't see how the asymptotic series not converging is a problem though: by definition the difference between the true value and the $k$th partial is $\sim$ to the $(k+1)$st term, which is exactly what we want in this case. – anon Apr 28 '22 at 01:26
  • Also we have $$\int_0^1\int_0^1\frac{(xy)^n}{1-xy},dx,dy=\frac{\pi^2}{6}-\sum_{k=1}^{n}\frac{1} {k^2}$$ and probably one can use this to evaluate the limit. – Paramanand Singh Apr 28 '22 at 03:40

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Let $N\geq 1$. Note that $$ \frac{{\pi ^2 }}{6} - \sum\limits_{n = 1}^N {\frac{1}{{n^2 }}} = \sum\limits_{n = N + 1}^\infty {\frac{1}{{n^2 }}} = \sum\limits_{n = N + 1}^\infty {\frac{1}{{n(n - 1)}}} - \sum\limits_{n = N + 1}^\infty {\frac{1}{{n^2 (n - 1)}}} = \frac{1}{N} - \sum\limits_{n = N + 1}^\infty {\frac{1}{{n^2 (n - 1)}}} . $$ Then $$ \sum\limits_{n = N + 1}^\infty {\frac{1}{{n^2 (n - 1)}}} \le \sum\limits_{n = N}^\infty {\frac{1}{{n^3 }}} \le \frac{1}{{N^{3/2} }}\sum\limits_{n = N}^\infty {\frac{1}{{n^{3/2} }}} \le \frac{{\zeta (3/2)}}{{N^{3/2} }}. $$ Accordingly, $$ \left( {\sum\limits_{n = 1}^N {\frac{1}{{n^2 }}} - \frac{{\pi ^2 }}{6}} \right)N = - 1 + \mathcal{O}\!\left( {\frac{1}{{N^{1/2} }}} \right), $$ showing that the limit is indeed $-1$.

Gary
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Following up on the suggestion of Paramanand Singh in the comments let $a_N =\sum_{n=1}^{N}\frac{1}{n^2}-\frac{\pi^2}{6}$ and $b_N=\frac{1}{N}$ Then $a_N \to 0$ and $b_N\to 0$ as $N \to \infty$ with $\{b_N\}$ strictly decreasing. We need to show $$ \lim _{N\to \infty}\frac{a_{N+1}-a_N}{b_{N+1}-b_N}= -1 $$ to deduce the lmit by applying the Cesaro-Stolz theorem, but $$ \begin{align} \frac{\frac{1}{(N+1)^2}}{\frac{1}{N+1}-\frac{1}{N}} &= \frac{\frac{1}{(N+1)^2}}{-\frac{1}{N(N+1)}}\\ \\ &= \frac{-N}{N+1}\to -1 \text{ as }N \to \infty \end{align} $$ Then by the Cesaro-Stolz theorem $\lim_{N\to\infty}\frac{a_N}{b_N}=-1$

sharding4
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Just for your curiosity

Making the problem more general $$S_N^{(k)}=N^{k-1} \Bigg[\sum_{n=1}^N \frac 1{n^k}-\sum_{n=1}^\infty \frac 1{n^k} \Bigg]=N^{k-1} \bigg[H_N^{(k)}-\zeta (k)\bigg]$$ Using the asymptotics of the generalized harmonic numbers $$H_N^{(k)}=N^{-k} \left(\frac{N}{1-k}+\frac{1}{2}-\frac{k}{12 N}+O\left(\frac{1}{N^3}\right)\right)+\zeta (k)$$ $$S_N^{(k)}=\frac{1}{1-k}+\frac{1}{2 N}-\frac{k}{12 N^2}+O\left(\frac{1}{N^4}\right)$$ which explains that for any $k$ you need to compute a lot of terms (just as you experienced).

Claude Leibovici
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Let $\epsilon>0$, there is a large enough $N$ such that $$\frac{1-\epsilon}{(N+m-1)(N+m)}<\frac1{(N+m)^2}<\frac1{(N+m)(N+m-1)}=\frac1{N+m-1}-\frac1{N+m}$$

for all $m\ge 0$.

Then we see that $$(1 - \varepsilon )(\frac{1}{{N - 1}} ) = (1 - \varepsilon )\sum\limits_m {(\frac{1}{{N + m - 1}} - \frac{1}{{N + m}})} < \sum\limits_m {\frac{1}{{{{(N + m)}^2}}} < \sum\limits_m {(\frac{1}{{N + m - 1}} - \frac{1}{{N + m}})} } = \frac{1}{{N - 1}}.$$

Then we can see the result easily.

Tianlalu
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stephenkk
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