Example: Try to calculate the series $\sum_{n=1}^{\infty} \frac{1}{n^{2 k}}$ for $k=1,2,3,4$.

 untranslated text remains the same as requested. However, the actual values for these series are well-known and can be provided as follows:

- For $k=1$, the series is $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$.
- For $k=2$, the series is $\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}$.
- For $k=3$, the series is $\sum_{n=1}^{\infty} \frac{1}{n^6} = \frac{\pi^6}{945}$.
- For $k=4$, the series is $\sum_{n=1}^{\infty} \frac{1}{n^8} = \frac{\pi^8}{9450}$.

These results are specific cases of the Riemann zeta function, $\zeta(s)$, where $s = 2k$.
We have (by (5))
$$\begin{array}{l}
\sum_{n=1}^{\infty} \frac{1}{n^{2}}=(-1)^{2} \frac{(2 \pi)^{2} B_{2}}{2 \cdot 2!}=\pi^{2} B_{2}=\pi^{2} / 6, \\
\sum_{n=1}^{x} \frac{1}{n^{4}}=(-1)^{3} \frac{(2 \pi)^{4} B_{4}}{2 \cdot 4!}=-\frac{\pi^{4}}{3} B_{4}=\pi^{4} / 90, \\
\sum_{n=1}^{\infty} \frac{1}{n^{6}}=(-1)^{4} \frac{(2 \pi)^{6} B_{6}}{2 \cdot 6!}=\frac{2 \pi^{6}}{45} B_{6}=\pi^{6} / 945, \\
\sum_{n=1}^{\infty} \frac{1}{n^{8}}=(-1)^{5} \frac{(2 \pi)^{8} B_{8}}{2 \cdot 8!}=-\frac{\pi^{8}}{315} B_{8}=\pi^{8} / 9450 .
\end{array}$$