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) $? 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?

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


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.
