\begin{align} \sum \limits_{n = 1}^{\infty}{\frac{1}{n^{2}}} &= \frac{\pi^2}{6} \\
\sum \limits_{n = 1}^{\infty}{\frac{1}{n^{4}}} &= \frac{\pi^4}{90} \\
\sum \limits_{n = 1}^{\infty}{\frac{1}{n^{6}}} &= \frac{\pi^6}{945} \\
\sum \limits_{n = 1}^{\infty}{\frac{1}{n^{8}}} &= \frac{\pi^8}{9450} \\
\sum \limits_{n = 1}^{\infty}{\frac{1}{n^{10}}} &= \frac{\pi^{10}}{93555} \\
\sum \limits_{n = 1}^{\infty}{\frac{1}{n^{12}}} &= \frac{691\pi^{12}}{638512875} \end{align}
- Für \( p \in \mathbb{N} \)
\begin{align} \sum \limits_{n = 1}^{\infty}{\frac{1}{n^{2p}}} &= { \frac{(-1)^{p+1}p\,\pi^{2p}}{(2p+1)!} – \sum \limits_{q = 1}^{p-1}\frac{(-1)^{p+q}\pi^{2p-2q}}{(2p-2q+1)!} \sum \limits_{n = 1}^{\infty}{\frac{1}{n^{2q}}}} \\
\frac{p\,\pi^{2p}}{(2p+1)!} &= \sum \limits_{q = 1}^{p}\frac{(-1)^{p+q}\pi^{2p-2q}}{(2p-2q+1)!} \sum \limits_{n = 1}^{\infty}{\frac{1}{n^{2q}}} \end{align}
\begin{align} \sum \limits_{n = 1}^{\infty}{\frac{1}{(2n-1)^{2p}}} &= \frac{2^{2p-1}-1}{2^{2p}}\sum \limits_{n = 1}^{\infty}{\frac{1}{n^{2p}}} \end{align}
\begin{align} \sum \limits_{n = 1}^{\infty}{\frac{(-1)^{n+1}}{(2n-1)^{3}}} &= \frac{\pi^{3}}{32} \\
\sum \limits_{n = 1}^{\infty}{\frac{(-1)^{n+1}}{(2n-1)^{5}}} &= \frac{5\pi^{5}}{1536} \\
\sum \limits_{n = 1}^{\infty}{\frac{1}{(2n-1)^{4}}} &= \frac{\pi^{4}}{96} \\
\sum \limits_{n = 1}^{\infty}{\frac{1}{(2n-1)^{6}}} &= \frac{\pi^{6}}{960} \end{align}
- Für \( r\in \mathbb{R},\, r\ne{0}\)
\begin{align} \sum \limits_{n = 1}^{\infty}{\frac{1}{1+r^2n^{2}}} &= \frac{1}{2} {\left[{\frac{\pi}{r}\coth{\left({\frac{\pi}{r}}\right)} – 1}\right]} \end{align}
\begin{align} \cdots &= \cdots \end{align}