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In math lesson, my teacher told me that Euler once used a very delicate method to calculate $\displaystyle S_{n}=\sum_{i=1}^{n}\frac{1}{i^{2}}$ and wrote a paper about it.

I wonder how to calculate this. I need a precise answer, not the answer that it's less than a number such as 2.

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marked as duplicate by J.R., Did, user127.0.0.1, user63181, Daniel Robert-Nicoud Feb 5 '14 at 12:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The finite sum or the series? $\endgroup$ – Martín-Blas Pérez Pinilla Feb 5 '14 at 11:05
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    $\begingroup$ If the sum was taken to be infinite, he would probably be refering to the paper "Remarques sur un beau rapport entre les series des puissances tant directes que reciproques" translated into english and open access in this pdf: rowan.edu/open/depts/math/osler/… $\endgroup$ – Graham Hesketh Feb 5 '14 at 11:15
  • $\begingroup$ This is Basel's Problem. Euler's method of finding the was a classic. $\endgroup$ – Inceptio Feb 5 '14 at 11:26
  • $\begingroup$ This question asks about the partial sum and has been closed on the grounds that the infinite sum has been tackled before; these are not the same thing... $\endgroup$ – Graham Hesketh Feb 5 '14 at 12:23
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For the infinite series: $$\sum _{n=1}^{\infty }\dfrac{1}{n^2}=\dfrac{{\pi }^{2}}{6} \tag{1}$$ you could see Euler's paper "Remarques sur un beau rapport entre les series des puissances tant directes que reciproques" translated into english and open access in this pdf. There are also many other nice derivations of that result on this website here. For the finite sum, note that as the series is absolutely convergent, we can write: $$ \begin{aligned} S_n=\sum _{i=1}^{n}\dfrac{1}{i^2}&=\sum _{i=1}^{\infty }\dfrac{1}{i^2}-\sum _{i=n+1}^{ \infty }\dfrac{1}{i^2}\\ &=\dfrac{{\pi }^{2}}{6}-\sum _{i=1}^{ \infty }\dfrac{1}{(i+n)^2} \end{aligned}\tag{2}$$ then, by the definition of the Polygamma fucntion: $${\frac {\Psi^{(m)} \left(n+1 \right) }{ \left( -1 \right) ^{m+1}m!}}= \sum _{i=1}^{\infty } \dfrac{1}{(i+n)^{m+1}} \tag{3}$$ (2) becomes: $$S_n=\dfrac{{\pi }^{2}}{6}-\Psi^{(2)} \left(n+1 \right) \tag{4}$$

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