How can we prove that the sum of alternating squares equals pi squared over 12?

[psholtz]

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

<br /> \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}<br />

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:

<br /> \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + ... = ?<br />

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?

 

[micromass]

Depends on what you mean with straightforward. Do you think Fourier series is straightforward? That allows you to prove it.

 

[Mute]

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

<br /> \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}<br />

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:

<br /> \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + ... = ?<br />

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).

 

[psholtz]

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?

 

[psholtz]

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..

 

[micromass]

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).

Oh, this works as well! Nice!

My idea was to work with the Fourier series of f(x)=x^2

 

[Mute]

Darn, guys, you replied too fast and quoted my post before I could edit in the missing word "make". =P