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In this answer two sequences are mentioned. In particular, I would like to prove that

$$\sum_{n = 1}^{+ \infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$

If I knew that the sequence converges to $\frac{\pi^2}{6}$, I could use the $\epsilon$-$M$ criterion to prove the convergence to that value.

But how to prove that the above sequence converges to that value if I don't know the value itself? Is there a general way to proceed in such cases?

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marked as duplicate by Bungo, Américo Tavares calculus Dec 9 '15 at 17:46

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    $\begingroup$ Euler gave a beautiful proof of this. Try to take a look. $\endgroup$ – Kushal Bhuyan Dec 9 '15 at 17:11
  • $\begingroup$ See This $\endgroup$ – Mark Viola Dec 9 '15 at 17:50
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You can evaluate the summation by evaluating the double integral $\displaystyle \int_{0}^1 \int_{0}^1 \dfrac{1}{1-xy}dx dy$ (it is an exercise to prove that this indeed equals $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2}$) and making the change of variables $x = \dfrac{u-v}{\sqrt{2}}$, $y= \dfrac{u+v}{ \sqrt{2}}$. This gives a rotation of $\dfrac{\pi}{4}$ about the origin through the angle $\dfrac{\pi}{4}$.

Although it is interesting to note that the exact value of $\displaystyle \int_{0}^1 \int_{0}^1 \int_{0}^1 \dfrac{1}{1-xyz}dx dy dz = \sum_{n = 1}^\infty \dfrac{1}{n^3}$ is unknown.

Note: This was not the way Euler solved the problem. He solved it more algebraically and used the Taylor series expansion for sine.

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  • $\begingroup$ It's worth mentioning that Euler's "proof" is not a rigorous argument. $\endgroup$ – Wojowu Dec 9 '15 at 17:32

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