Infinite series for these two values.

Question

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:

$$ \dfrac{3 \pi^3 \sqrt 2}{16} = \sum \text{Some expression}. $$

Answer 1

$\displaystyle \sum_{n=1}^\infty\frac1{(2n-1)^2(2n+1)^2}=\frac{\pi^2-8}{16}$ (see this).

$\displaystyle 8\sum_{n=-\infty}^\infty \frac{(-1)^n}{(4n+1)^3} = \frac{3\pi^3 \sqrt{2}}{16}$ (see this).

Answer 2

For any real number $a$ and any integer $b$, we have

$$a = \sum\limits_{n=0}^\infty \dfrac{\lfloor b^n a\rfloor - b\lfloor b^{n-1} a\rfloor}{b^n}.$$

Answer 3

I think my answer will be trivial as compared to other's but still with my high-school knowledge , I can answer your first part-

$(6/16)\sum_{n=0}^{\infty} ((1/n^2-4/(3.2^n))= (\pi^2-8)/16 $