Question

\[\sum _{n=1}^{\infty } \frac{1}{n^2}\]

What's the Sum

Answer 1

F(x)=|x|
from -L to L

May be represented using Fourier series as
\[|x|=\frac{L}{2}- \frac{(4L)}{\pi ^2}\sum _{n=1,3,5}^{\infty } \frac{1}{n^2}\text{Cos}\left[\frac{\text{n$\pi $} x}{L}\right]\]

\[F[0]=\frac{L}{2}- \frac{(4L)}{\pi ^2}\sum _{n=1,3,5}^{\infty } \frac{1}{n^2}\]


\[0=\frac{1}{2}- \frac{(4)}{\pi ^2}\sum _{n=1,3,5}^{\infty } \frac{1}{n^2}\]


\[\frac{1}{2}= \frac{(4)}{\pi ^2}\sum _{n=1,3,5}^{\infty } \frac{1}{n^2}\]

\[\frac{\pi ^2}{8}=\left(1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\text{...}.\right)\]

\[\sum _{n=1}^{\infty } \frac{1}{n^2}=\left(1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\text{...}.\right)+\left(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\text{...}.\right)\]

\[\sum _{n=1}^{\infty } \frac{1}{n^2}=\left(1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\text{...}.\right)+\frac{1}{4}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\text{...}.\right)\]

\[\frac{3}{4}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\text{...}.\right)=\frac{\pi ^2}{8}\]

\[\frac{3}{4}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\text{...}.\right)=\frac{\pi ^2}{6}\]

Answer 2

Yes. That's one proof, and a nice one too. This is famous problem and there quite a few proofs floating around.

Answer 3

Is there a Proof which can be understood by me?

Answer 4

I think Euler did something to proof this as well?

Answer 5

Well it's just algebra after the fourier series part

Answer 6

I don't have a Clue about the fourier series
I always end up readin furier as fuhrer

Answer 7

Oh I see, I think maybe I got it; little bit of fourier series.

Answer 8

What gave you the intuition to use Fourier Series? Or is that just somewhat of a standard trick for this series?