Prove that $\sum \frac{1}{n^2} = \frac{\pi^2}{6}$ [duplicate]

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?

• Euler gave a beautiful proof of this. Try to take a look. – Kushal Bhuyan Dec 9 '15 at 17:11
• See This – Mark Viola Dec 9 '15 at 17:50

1 Answer

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.

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