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What will be the value of $$\sum_{r=1}^n \frac{1}{r^2}$$ in terms of finite $n$?

I tried to solve it using $V_n$ method but couldn't get how to convert this series in telscopic series.

Please anyone help me to solve it.

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Sameer Nilkhan
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    Does this answer your question? Why does the sum of inverse squares equal $\pi^2/6$? – Matti P. Dec 31 '19 at 09:25
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    The series is well-known to converge. But there is no simple expression for its partial sums. Some convergent series (like $1+\frac{1}{2}+\frac{1}{2^2}+\dots$) have simple expressions for the partial sums, but most don't. – almagest Dec 31 '19 at 09:32
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    The question is inconsistent in its current form. A series is by definition the limit of the partial sums of a sequence. Thus, by referring to a "series", you seem to imply that you're interested in the limit of the partial sums (hence the above suggestion of a duplicate), whereas the rest of the question seems to suggest that you're actually interested in the partial sums themselves, not just the limit. Please clarify. – joriki Dec 31 '19 at 09:38
  • Yes i am interested in partial sums not limit. – Sameer Nilkhan Dec 31 '19 at 09:41
  • https://www.wolframalpha.com/input/?i=Sum%5B1%2Fr%5E2%2C%7Br%2C1%2Cn%7D%5D

    This sum defines a generalised harmonic number, and can be written in terms of the first derivative of the digamma function (sometimes called trigamma).

     – Benedict W. J. Irwin Dec 31 '19 at 09:54

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Using the generalized harmonic numbers$$S_n=\sum_{i=1}^n \frac 1 {i^2}=H_n^{(2)}$$ If $n$ is large, we can use the expansion $$S_n=\frac{\pi ^2}{6}-\frac{1}{n}+\frac{1}{2 n^2}-\frac{1}{6 n^3}+\frac{1}{2 n^4}-\frac{29}{30 n^5}+\frac{3}{2 n^6}+O\left(\frac{1}{n^7}\right)$$ For example, $$S_{10}=\frac{1968329}{1270080}\approx 1.54977$$ while the above truncated series gives $$S_{10} \sim \frac{\pi ^2}{6}-\frac{570749}{6000000}\approx 1.54981$$

Claude Leibovici
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This is a bit backwards, but might help you work out what is required:

If we can write $$ \sum_{r=1}^n \frac{1}{r^2} = \frac{\pi^2}{6} - \psi^{(1)}(n+1) = H^{(2)}_n $$ we could plausibly guess that this involves $$ \sum_{r=1}^\infty \frac{1}{r^2} = \frac{\pi^2}{6} $$ where then all of the higher terms are subtracted off, according to Wikipedia $$ \psi^{(1)}(n+1) = \sum_{r=0}^\infty \frac{1}{(n+1+r)^2} $$ which might be the "telescoping" you were referring to?

Benedict W. J. Irwin
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