How can we see that $ \sum_{n=0}^{\infty}\frac{2^n(1-n)^3}{(n+1)(2n+1){2n \choose n}}=(\pi-1)(\pi-3) $?

Question

I wonder will it help me so prove it if I was to decompose it into partial fractions?

Mathematica approves of the identity; it is converges. can anyone help me to prove it?

$$ \sum_{n=0}^{\infty}\frac{2^n(1-n)^3}{(n+1)(2n+1){2n \choose n}}=(\pi-1)(\pi-3) $$

Answer 1

Hint. One may observe that, for $n\geq 0$, using the Euler beta function, $$ \frac{2^n(1-n)^3}{(n+1)(2n+1){2n \choose n}}=-\frac{(n-1)^3}{n+1}\int_0^1(2x(1-x))^ndx. \tag1 $$ Then, one may write

$$\begin{align} &\sum_0^\infty\frac{2^n(1-n)^3}{(n+1)(2n+1){2n \choose n}} \\\\&=-\sum_0^\infty\frac{(n-1)^3}{n+1}\int_0^1(2x(1-x))^ndx\\\\ &=-\int_0^1\sum_0^\infty\frac{(n-1)^3}{n+1}(2x(1-x))^n \:dx\\\\ &=\int_0^1\left(\frac{-7+34 x-82 x^2+96 x^3-48 x^4}{(1-2 x+2 x^2)^3}-\frac{4 \log(1-2 x+2 x^2)}{(1-x) x}\right)dx\\\\ &=\underbrace{\int_0^1\frac{-7+34 x-82 x^2+96 x^3-48 x^4}{(1-2 x+2 x^2)^3}\,dx}_{\large \color{blue}{3-4\pi}}+\underbrace{4\int_0^1-\frac{\log(1-2 x+2 x^2)}{(1-x) x}\,dx}_{\large \color{red}{\pi^2}} \\\\&=\color{blue}{3-4\pi}+\color{red}{\pi^2} \\\\&=(\pi-1)(\pi-3) \end{align} $$

as announced.