Does anyone know what $\displaystyle \sum_{n=1}^{\infty} \frac 1 {n^4}$ is? I think that it's $\displaystyle \frac{\pi^4} {90}$.
Wolfram|Alpha. That inequality is very well known by the way.
Haha, sorry I was just doing inequalities. So, how exactly did you work it out?
P.S. there are formulas for the values for arbitrary powers.
P.P.S. http://www.wolframalpha.com/input/?i...inity+of+1/n^4
That's cool and understandable.
I was thinking maybe you used $\displaystyle \frac{cot({\pi}z)}{z^{4}}$
Did you use a rectangle centered at the origin with $\displaystyle n+\frac{1}{2}, \;\ n-\frac{1}{2}$ along the real axis and $\displaystyle -ni, \;\ ni$ along the imaginary axis?.
I have seen it donw this way with $\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^{2}}$
The residue at 0 is
$\displaystyle \frac{-{\pi}^{3}}{45}$
and $\displaystyle \frac{1}{{\pi}k^{4}}$ at $\displaystyle k\neq 0$
I played around a little and combined the residues at -k and k into:
$\displaystyle 2{\pi}i\cdot\frac{-{\pi}^{3}}{45}+4i\left(1+\frac{1}{2^{4}}+\frac{1}{ 3^{4}}+....\right)=0$
$\displaystyle \left(1+\frac{1}{2^{4}}+\frac{1}{3^{4}}+....\right )=\frac{{\pi}^{4}}{90}$
Like I said, just playing around. CA is not my forte, but I enjoy it. It is a powerful tool. What I have learned is self-taught, so if there is a booboo I am here to learn as well as help.
Another way to find the general expression for $\displaystyle \zeta(2n)$ is to expand $\displaystyle \pi \cot \pi z$ in a Laurent series about 0 using the usual series expansion (whose coefficients are the Bernoulli numbers). Then expand it in another way using the identity $\displaystyle \pi \cot \pi z = \frac{1}{z} + \sum_{j=1}^\infty\frac{1}{z^2-j^2} = \frac{1}{z} + \sum_{j=1}^\infty\frac{-1}{j^2}\frac{1}{1-z^2/j^2} = \frac{1}{z} + \sum_{j=1}^\infty\frac{-1}{j^2}\sum_{k=0}^\infty (z^2/j^2)^k$. Switching the sums and gathering like powers, we obtain $\displaystyle \zeta(2n)=\frac{(-1)^{n+1}(2\pi)^{2n}B_{2n}}{2(2n)!}$ after comparing coefficients in both expansions.