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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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    If you are interested in questions like these, Euler McLaurin ought to be in your toolbox!2012-03-30

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

Which according to my previous answer here: https://math.stackexchange.com/a/14518/1102 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: http://math.stackexchange.com/questions/14515/what-is-the-expression-of-n-that-equals-to-sum-i-1n-frac1i22012-03-30
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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.