Rick16 said:

So apparently there is no straight forward algorithm for this. I will look for an online calculator then.

In another thread I provide a derivation of the general closed formula for ##\zeta(2k)## in terms of Bernoulli numbers:

\begin{align*}
\zeta(2k) = \sum_{n=1}^\infty \frac{1}{n^{2k}} = (-1)^{k+1} (2 \pi)^{2k} \dfrac{B_{2k}}{2 (2k)!}
\end{align*}

From which you can obatin your series as explained by @pasmith in this other thread. The thread is found here: Question on an infinite summation series. This derivation, contained in the attachment, uses complex contour integration directly.

In a second method, presented in the thread itself, relates the infinite sum to a real integral, which is then evaluated using contour integration to obtain:

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^{2k}} & = (1-2^{-2k}) (-1)^{k+1} (2 \pi)^{2k} \frac{B_{2k}}{2 (2k)!} .
\end{align*}

In the second method, I use an identity that can be derived more directly than in the derivation given in the thread:

\begin{align*}
\int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x - 1)^2} dx - \int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x + 1)^2} & = \int_{-\infty}^\infty \dfrac{4 x^{2k} e^{2x}}{(e^{2x} - 1)^2} dx
\nonumber \\
& = \int_{-\infty}^\infty \dfrac{2^{-2k+1} (2x)^{2k} e^{2x}}{(e^{2x} - 1)^2} 2dx
\nonumber \\
& = 2^{-2k+1} \int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x - 1)^2} dx
\end{align*}

implies

\begin{align*}
\int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x - 1)^2} dx = \frac{1}{1- 2^{-2k+1}} \int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x + 1)^2}
\end{align*}

Extracting relevant bits from other thread: We can write

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} & = \sum_{n=1}^\infty \dfrac{1}{n^2} - \sum_{n=1}^\infty \dfrac{1}{(2n)^2}
\nonumber \\
& = (1 - 2^{-2}) \sum_{n=1}^\infty \dfrac{1}{n^2}
\nonumber \\
& = \frac34 \sum_{n=1}^\infty \dfrac{1}{n^2}
\end{align*}

We have

\begin{align*}
\sum_{n=1}^\infty \dfrac{1}{n^2} & = \sum_{n=1}^\infty \dfrac{1}{n^2} \int_0^\infty e^{-y} y dy
\nonumber \\
& = \sum_{n=1}^\infty \int_0^\infty e^{-nx} x dx
\nonumber \\
& = \int_0^\infty \dfrac{x}{e^{x} - 1} dx
\nonumber \\
& = \frac{1}{2} \int_0^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx
\nonumber \\
& = \frac{1}{4} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx
\end{align*}

So

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} & = \frac{3}{16} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx \quad (*)
\end{align*}

Note that

\begin{align*}
\int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx - \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} & = \int_{-\infty}^\infty \dfrac{4 x^2 e^{2x}}{(e^{2x} - 1)^2} dx
\nonumber \\
& = \int_{-\infty}^\infty \dfrac{2^{-2+1} (2x)^2 e^{2x}}{(e^{2x} - 1)^2} 2dx
\nonumber \\
& = 2^{-2+1} \int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x - 1)^2} dx
\end{align*}

implies

\begin{align*}
\int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx = 2 \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2}
\end{align*}

Substituting this into ##(*)##,

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} & = \frac38 \int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x + 1)^2} dx
\end{align*}

Consider the integral:

\begin{align*}
\int_{-\infty}^\infty \dfrac{e^{\alpha x} e^x}{(e^{x} + 1)^2} dx
\end{align*}

where ##-\frac{1}{2} \leq \alpha \leq \frac{1}{2}##. Then

\begin{align*}
\int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx = \left. \dfrac{\partial^2}{\partial \alpha^2} \int_{-\infty}^\infty \dfrac{e^{\alpha x} e^x}{(e^{x} + 1)^2} dx \right|_{\alpha=0}
\end{align*}

You can evaluate this integral using complex analysis by considering the rectangular contour, ##C##, in the figure

and the integral

\begin{align*}
\oint_C \dfrac{e^{\alpha z} e^z}{(e^z + 1)^2} dz
\end{align*}

whose integrand has a pole at ##\pi i##.