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What is the expression of $n$ that equals to $\sum_{i=1}^n \frac{1}{i^2}$?

Asymptotic formulas for the n-th harmonic number are well-known: $$ \sum_{k=1}^n\frac1n=\log n+\gamma+\frac{1}{2n}-\frac{1}{12n^2}+O\left(\frac{1}{n^4}\right) $$ with more terms easy to generate if needed.

Is there is a similar formula for the sums of reciprocal squares? Something like $$ \sum_{k=1}^n\frac{1}{n^2}=\frac{\pi^2}{6}+\cdots+O\left(\frac{1}{n^4}\right) $$ (order negotiable, but ideally something decent like the above).

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marked as duplicate by Aryabhata, robjohn, Dylan Moreland, t.b., anon Mar 31 '12 at 0:08

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.

If you are interested in questions like these, Euler McLaurin ought to be in your toolbox! –  Aryabhata Mar 30 '12 at 23:14

2 Answers 2

up vote 2 down vote accepted

Yes, I believe you can get this from the Euler McLaurin Summation formula.

Which according to my previous answer here: comes to (if you only include upto $n^{-2}$)

$$\sum_{j=1}^{n} \dfrac{1}{j^2} = \frac{\pi^2}{6} - \dfrac{1}{n} - \dfrac{1}{2n^2} + \mathcal{O}(\dfrac{1}{n^3})$$

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Wait.. this is a dupe... Yes:… –  Aryabhata Mar 30 '12 at 23:04

Wikipedia gives the identity

$$\int_0^a H_{x,2}\;\mathrm dx=a\frac{\pi^2}6-H_a$$

where $H_{n,m}=\sum_{k=0}^n k^{-m}$ and $H_n = H_{n,1}$. It may be useful to get an approximation.

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