13

Basel series $$ \lim_{n\to \infty} \sum_{k=1}^{n} \frac{1}{k^2} = \frac{{\pi}^2}{6} $$
is well known. I'm interested in computing the limit value $$ \lim_{n \to \infty} n\left(\frac{{\pi}^2}{6} - \sum_{k=1}^{n} \frac{1}{k^2} \right) $$

although I am not sure even whether this limit exists or not..

Does this limit exists? If exists, how can we compute this?

Thanks in advance.

Guldam
  • 934

3 Answers3

12

By integral test,

$$\int_{n+1} ^\infty \frac{1}{x^2} dx \le \sum_{k = n+1}^\infty \frac{1}{k^2} \le \int_n ^\infty \frac{1}{x^2} dx$$

Thus

$$\frac{1}{n+1} \le \sum_{k = n+1}^\infty \frac{1}{k^2} \le \frac{1}{n}. $$

5

Hint

By definition, $$\sum_{k=1}^{n} \frac{1}{k^2} = H_n^{(2)}$$ where appears the harmonic number. For large values of $n$, $$H_n^{(2)}=\frac{\pi ^2}{6}-\frac{1}{n}+\frac{1}{2 n^2}+O\left(\left(\frac{1}{n}\right)^3\right)$$ Then, the result.

Claude Leibovici
  • 260,315
  • how exactly did you get that precise estimate? – happymath Oct 09 '15 at 07:51
  • 1
    This nice answer would be a very good answer if you mentioned (just by name) a method of obtaining that asymptotics, and additionally explained that one could, in principle, obtain as many terms as one wish, because that would open the door to the nice area of asymptotics to OP... – mickep Oct 09 '15 at 07:55
  • 1
    @happymath. It is a classical expansion of the harmonic numbers. You can generalize it as $$H_n^{(m)}=n^{-m} \left(-\frac{n}{m-1}+\frac{1}{2}-\frac{m}{12 n}+O\left(\left(\frac{1}{n}\right)^3\right)\right)+\zeta (m)$$ – Claude Leibovici Oct 09 '15 at 07:56
  • @ClaudeLeibovici thanks for the general formula. Can you please give a reference for the result? – happymath Oct 09 '15 at 08:01
  • @happymath. I don(t have any reference for that. You can get any of these from series expansion for large values of $n$. It is something I worked decades ago. – Claude Leibovici Oct 09 '15 at 08:09
  • 1
    Here are some refences and a few more terms: With the Digamma function you have from http://functions.wolfram.com/GammaBetaErf/HarmonicNumber2/03/01/01/0009/ $$\sum_{k=1}^{n} \frac{1}{k^2} = H_n^{(2)} = \psi^{(1)}(1)-\psi^{(1)}(n+1)$$ You can either use the somewhat complicated asymptotic expansion from http://functions.wolfram.com/GammaBetaErf/HarmonicNumber2/06/02/0001/ or calculate it with e.g. Maple $$H_n^{(2)} \sim \frac{1}{6}{\pi^2} - \frac{1}{n} + \frac{1}{2}\frac{1}{n^2}-\frac{1}{6}\frac{1}{n^3} +\frac{1}{30}\frac{1}{n^5}+O(\frac{1}{n^6})$$ – gammatester Oct 09 '15 at 09:37
  • @gammatester. Thank you very much ! – Claude Leibovici Oct 09 '15 at 09:47
2

By the Stolz rule, for $a_n =\frac{\pi^2}{6} - \sum\limits_{k=1}^n \frac{1}{k^2}$ and $b_n = \frac{1}{n}\downarrow 0$, $$ \lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{a_{n-1}-a_n}{b_{n-1}-b_n} = \lim_{n\to\infty} \frac{\frac{1}{n^2}}{\frac{1}{n-1}-\frac{1}{n}} = 1. $$

This also can be used to get further asymptotics, e.g. $$ \lim_{n\to\infty} \frac{\frac{\pi^2}{6} - \sum\limits_{k=1}^n \frac{1}{k^2} - \frac{1}{n}}{\frac{1}{n^2}} = \lim_{n\to\infty} \frac{\frac{1}{n^2} - \frac{1}{n(n-1)}}{\frac{1}{(n-1)^2} - \frac{1}{n^2}} = \frac{1}{2} $$ et cetera.

zhoraster
  • 25,481