Convergence to pi^2/6

[bomba923]

How does
infinity
Sum (n^(-2)=(pi^2)/6
n=1

Please tell me if this has been posted before (afraid :redface: )
(in that case, i'll see the other post)

[master_coda]

Take f(x)=x. Then the Fourier coeffcients of f are a_n=0 and b_n=\frac{2}{n}(-1)^{n+1}. Parseval's theorem says that:

\frac{1}{\pi}\int_{-\pi}^\pi f^2(x)\:dx=\frac{a_0^2}{2}+\sum_{k=1}^\infty\left( a_k^2+b_k^2)

Since the a_n terms are all zero, this reduces to:

\frac{1}{\pi}\int_{-\pi}^\pi x^2\:dx=4\sum_{n=1}^\infty\frac{1}{n^2}

The integral is easy enough to solve, and the left hand side reduces to 2\pi^2/3. Dividing both sides by four gives us:

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

[bomba923]

I see clearly now!--thanks

(Will no one answer my "digit-factorial question" thread :frown: )

[dextercioby]

And let's not forget Euler's original method.Combining the series he found for \frac{\sin x}{x} and the one from Taylor expansion,he was able to prove it...

Daniel.

[matt grime]

And there's another way:

\int\int\frac{1}{1-xy}dxdy

Evalutate that as x and y both go from 0 to 1. Do it using a substitution, and then do it by replacing the fraction inside with its series expansion and ignore the convergence issues to rearrange sum and integral.

[SpaceDomain]

So this is parsevals theorem?

\frac{1}{\pi}\int_{-\pi}^\pi f^2(x)\:dx=\frac{a_0^2}{2}+\sum_{k=1}^\infty\left( a_k^2+b_k^2)

?

[Mark44]

So this is parsevals theorem?

\frac{1}{\pi}\int_{-\pi}^\pi f^2(x)\:dx=\frac{a_0^2}{2}+\sum_{k=1}^\infty\left( a_k^2+b_k^2)

?
How about this?

\frac{1}{\pi}\int_{-\pi}^\pi f^2(x)\:dx=\frac{a_0^2}{2}+\sum_{k=1}^\infty( a_k^2+b_k^2)