I know that if I want a infinite series whose result is $4$, I could do something like this:
$$
4= \dfrac{1}{1/4} = \dfrac{1}{\dfrac{4-3}{4}} = \dfrac{1}{1 -\dfrac{3}{4}} = \sum_{n=0}^{\infty} \left(\dfrac{3}{4} \right)^n,
$$
using the famous relation:
$$
\dfrac{1}{1-x} = \sum_{n=0}^{\infty} x^n, \quad |x| < 1.
$$
I think this works for every rational number. Another example is a infinite series whose result is $\sqrt 2$, as shown by this answer:
$$
\begin{equation*}
\sqrt2 =\sum_{k=0}^\infty\frac{(2k-1)!!}{4^kk!}.
\end{equation*}
$$
There are some infinite series involving $\pi$ such as:
$$
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^2} &= \dfrac{\pi^2}{6},\\
\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{2n-1} &= \dfrac{\pi}{4}, \\
\sum_{n=1}^{\infty} \dfrac{1}{n^4} &= \dfrac{\pi^4}{90}, \\
\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{(2 n-1)^3} = &= \dfrac{\pi^3}{32}.
\end{align}
$$
Just like this, I would like to know if there is a infinite series with rational terms for:
$$
\dfrac{\pi^2 – 8}{16} = \sum \text{Some expression}
$$
and for:
$$
\dfrac{3 \pi^3 \sqrt 2}{16} = \sum \text{Some expression}.
$$
Best Answer
$\displaystyle \sum_{n=1}^\infty\frac1{(2n-1)^2(2n+1)^2}=\frac{\pi^2-8}{16}$ (see this).
$\displaystyle 8\sum_{n=-\infty}^\infty \frac{(-1)^n}{(4n+1)^3} = \frac{3\pi^3 \sqrt{2}}{16}$ (see this).