# Calculating Riemann Zeta function

1. Dec 8, 2008

### Bill Foster

1. The problem statement, all variables and given/known data

Using method of Euler, calculate $$\zeta(4)$$, the Riemann Zeta function of 4th order.

2. Relevant equations

$$\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}$$

Finding $$\zeta(2)$$:

$$\zeta(2)=\sum_{n=1}^\infty \frac{1}{n^s}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...$$

$$\sin{x}=x-\frac{1}{3!}x^3+\frac{1}{5!}x^5+...$$
$$\frac{\sin{x}}{x}=1-\frac{1}{3!}x^2+\frac{1}{5!}x^4+...$$

$$\frac{\sin{x}}{x}=\left(1-\left(\frac{x}{\pi}\right)^2\right)\left(1-\left(\frac{x}{2\pi}\right)^2\right)\left(1-\left(\frac{x}{3\pi}\right)^2\right)\left(1-\left(\frac{x}{4\pi}\right)^2\right)\left(1-\left(\frac{x}{5\pi}\right)^2\right)...$$

From the above, the coefficients of $$x^2$$ are:

$$\frac{1}{\pi^2}+\frac{1}{2^2\pi^2}+\frac{1}{3^2\pi^2}+\frac{1}{4^2\pi^2}+\frac{1}{5^2\pi^2}+...$$

Now equate these coefficients to $$x^2$$ in the sine function series:

$$\frac{1}{\pi^2}+\frac{1}{2^2\pi^2}+\frac{1}{3^2\pi^2}+\frac{1}{4^2\pi^2}+\frac{1}{5^2\pi^2}+...=\frac{1}{3!}$$
$$\frac{1}{1}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...=\frac{\pi^2}{3!}$$
$$\zeta(2)=\frac{\pi^2}{6}$$

3. The attempt at a solution

I get the following for the coefficients of $$x^4$$:

$$\frac{1}{2^2\pi^4}+\frac{1}{3^2\pi^4}+\frac{1}{2^23^2\pi^4}+\frac{1}{4^2\pi^4}+\frac{1}{2^24^2\pi^4}+\frac{1}{3^24^2\pi^4}+...=\frac{1}{\pi^4}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{2^23^2}+\frac{1}{4^2}+\frac{1}{2^24^2}+\frac{1}{3^24^2}+...\right)$$

The problem is, how do I get $$\zeta(4)=\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+...$$ out of that sum?

Last edited: Dec 8, 2008