# Showing $\sum _{k=1} 1/k^2 = \pi^2/6$ [duplicate]

I read my book of EDP, and there appears the next serie $$\sum _{k=1} \dfrac{1}{k^2} = \dfrac{\pi^2}{6}$$ And, also, we prove that this series is equal $\frac{\pi^2}{6}$ for methods od analysis of Fourier, but...

Do you know other proof, any more simple or beautiful?

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## marked as duplicate by David Mitra, Henry, Asaf Karagila, Martin Sleziak, Nate EldredgeMay 13 '12 at 20:35

That should be $\pi^2/6$. – David Mitra May 13 '12 at 19:31
See here – David Mitra May 13 '12 at 19:33
@Martin: This question asks for methods to calculate the sum, not to prove its convergence. This is at least how I read this question. – Asaf Karagila May 13 '12 at 19:55
@Asaf I think you' right. Although, the body is different from the title. Which explains why I thought that the OP asks about convergence only. (It's not that important now, since we found duplicates for both possible meanings.) – Martin Sleziak May 13 '12 at 19:57
The title originally said Prove that this series converges – Henry May 14 '12 at 10:05

If you just want to show it converges, then the partial sums are increasing but the whole series is bounded above by $$1+\int_1^\infty \frac{1}{x^2} dx=2$$ and below by $$\int_1^\infty \frac{1}{x^2} dx=1,$$ since $\int_{k}^{k+1} \frac{1}{x^2} dx \lt \frac{1}{k^2} \lt \int_{k-1}^{k} \frac{1}{x^2} dx$.