Can Euler's Method Solve Higher Order Sums in the Basel Problem?

[Parmenides]

Hello,

In several courses now, I have seen the following:

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

and

$$\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{{\pi}^4}{90}$$

and so forth. While I know that these are related to the Riemann Zeta Function for even powers of n, I was wondering if there was a way to analytically solve these sorts of sums, without recourse to it. Is it possible to extend Euler's method of solving the Basel Problem to higher orders of even n (expanding the sine series, collecting the roots, and equating terms)?

 

[micromass]

Euler's argument was very nonrigorous. It worked, but he failed to prove most of the argument.

That said, there are analytic solutions of the series you mention and there are rigorous proofs. One of those proofs is through complex analysis and the residue theorem. Another proofs is through Fourier analysis. I'm sure there are many others. The analytic solution of your series can be found on wikipedia: http://en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values

So,

$$\sum_{n=1}^{+\infty}\frac{1}{n^s}= (-1)^{s+1}\frac{B_{2s}(2\pi)^{2s}}{2(2s)!}$$

For the proof, I will have to refer you to textbooks. I know "Complex Analysis" by Freitag and Busam is an excellent text which contains the proof as an application of the residue theorem. I'm sure others can give other good references.

 

[jbunniii]

This result is also proved without any complex analysis in Courant and John, Introduction to Calculus and Analysis, Vol. I, in the chapter on trigonometric series.