## Basel Problem

The Basel Problem is a well known result in analysis which basically states:

$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}$$

There are various well-known ways to prove this.

I was wondering if there is a similar, simple way to calculate the value of the sum:

$$\frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + ... = ???$$

The value of this sum should work out to pi*pi/12, but I was wondering if there was a straightforward way to prove it?

 PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Depends on what you mean with straightforward. Do you think Fourier series is straightforward? That allows you to prove it.

Recognitions:
Homework Help
 Quote by psholtz The Basel Problem is a well known result in analysis which basically states: $$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}$$ There are various well-known ways to prove this. I was wondering if there is a similar, simple way to calculate the value of the sum: $$\frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + ... = ???$$ The value of this sum should work out to pi*pi/12, but I was wondering if there was a straightforward way to prove it?
Well, it may not be rigorous, but you could write

$$\begin{eqnarray*} \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + ...& = & \left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots\right) - 2\left(\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \dots \right) \\ & = & \left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots\right) - \frac{2}{2^2}\left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots \right) \\ & = & \dots\end{eqnarray*}$$

(You could make this rigourous by being careful with the convergence of the two series, I imagine).

## Basel Problem

Sure, a Fourier series would be straightforward.

I'm familiar w/ how Fourier analysis can be used to sum the first series, but it's not immediately clear to me how to proceed from that solution, to the sum for the second series.

Could you give me a pointer/hint?

 Quote by Mute Well, it may not be rigorous, but you could write $$\begin{eqnarray*} \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + ...& = & \left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots\right) - 2\left(\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \dots \right) \\ & = & \left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots\right) - \frac{2}{2^2}\left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots \right) \\ & = & \dots\end{eqnarray*}$$ (You could this rigourous by being careful with the convergence of the two series, I imagine).
Yes, thanks...

This was something along the lines of the intuition I was going by, but didn't quite get it to this point.

Thanks..

Blog Entries: 8
Recognitions:
Gold Member
 Quote by Mute Well, it may not be rigorous, but you could write $$\begin{eqnarray*} \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + ...& = & \left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots\right) - 2\left(\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \dots \right) \\ & = & \left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots\right) - \frac{2}{2^2}\left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots \right) \\ & = & \dots\end{eqnarray*}$$ (You could this rigourous by being careful with the convergence of the two series, I imagine).
My idea was to work with the Fourier series of $f(x)=x^2$