Showing $\sum _{k=1} 1/k^2 = \pi^2/6$

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Different methods to compute $\sum\limits_{n=1}^\infty \frac{1}{n^2}$
Does $\sum\limits_{k=1}^n 1 / k ^ 2$ converge when $n\rightarrow\infty$?

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?

@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

score 15 · Answer 1 · edited May 13 '12 at 20:23

15

Fourteen proofs compiled by Robin Chapman.

http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf

edited May 13 '12 at 20:23

answered May 13 '12 at 19:56

Gaston Burrull

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score 4 · Answer 2 · answered May 13 '12 at 19:44

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 $\int_{k}^{k+1} \frac{1}{x^2} dx \lt \frac{1}{k^2} \lt \int_{k-1}^{k} \frac{1}{x^2} dx$.

Showing $\sum _{k=1} 1/k^2 = \pi^2/6$

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