TITLE: How can we show that $ \sum_{n=0}^{\infty}\frac{2^nn[n(\pi^3+1)+\pi^2](n^2+n-1)}{(2n+1)(2n+3){2n \choose n}}=1+\pi+\pi^2+\pi^3+\pi^4 ?$
QUESTION [13 upvotes]: We proposed this sum, but we are lacking in knowledge of this area of maths and we would ask if any of the authors would be willing to show us step by step how to go about proving this sum.
$$
\sum_{n=0}^{\infty}\frac{2^nn[n(\pi^3+1)+\pi^2](n^2+n-1)}{(2n+1)(2n+3){2n \choose n}}=1+\pi+\pi^2+\pi^3+\pi^4
$$

REPLY [3 votes]: Here is an answer based upon the arcsine function.

We start with the following formula valid for $u\in(0,2)$
  \begin{align*}
A(u)&=\sum_{n=0}^\infty\frac{2^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}u^n\\
&=\frac{1}{u}-\frac{1}{u}\sqrt{\frac{2}{u}-1}\cdot\arcsin\left(\sqrt{\frac{u}{2}}\right)\\
&=\frac{1}{6}+\frac{1}{30}u+\frac{1}{105}u^2+\frac{1}{315}u^3+\frac{4}{3465}u^4+\frac{4}{9009}u^5+\cdots\\
\end{align*}
A rather detailed derivation  can be found in this answer.

$$ $$

Building blocks: $(uD_u)^kA(u)$
The next step is to successively apply the operator $uD_u$ on $A(u)$ with $D_u$ the differential operator. We obtain with some help of Wolfram Alpha
  \begin{align*}\
(uD_u)A(u)&=\sum_{n=0}^\infty\frac{n2^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}u^n\\
&=-\frac{3}{2u}+\frac{3-u}{u^2\sqrt{\frac{2}{u}-1}}\cdot\arcsin\left(\sqrt{\frac{u}{2}}\right)\\
&=\frac{1}{30}u+\frac{2}{105}u^2+\frac{1}{105}u^3+\frac{16}{3465}u^4+\frac{20}{9009}u^5+\cdots\\
\\
(uD_u)^2A(u)&=\sum_{n=0}^\infty\frac{n^22^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}u^n\\
&=\frac{4u-9}{2u(u-2)}+\frac{u^2-7u+9}{u^2(u-2)\sqrt{\frac{2}{u}-1}}\cdot\arcsin\left(\sqrt{\frac{u}{2}}\right)\\
&=\frac{1}{30}u+\frac{4}{105}u^2+\frac{1}{35}u^3+\frac{64}{3465}u^4+\frac{100}{9009}u^5+\cdots\\
\\
(uD_u)^3A(u)&=\sum_{n=0}^\infty\frac{n^32^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}u^n\\
&=-\frac{5u^2-25u+27}{2u(u-2)^2}-\frac{u^3-13u^2+34u-27}{u^2(u-2)^2\sqrt{\frac{2}{u}-1}}\cdot\arcsin\left(\sqrt{\frac{u}{2}}\right)\\
&=\frac{1}{30}u+\frac{8}{105}u^2+\frac{3}{35}u^3+\frac{256}{3465}u^4+\frac{500}{9009}u^5+\cdots\\
\\
(uD_u)^4A(u)&=\sum_{n=0}^\infty\frac{n^42^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}u^n\\
&=\frac{6u^3-53u^2+115u-81}{2u(u-2)^3}\\
&\qquad+\frac{u^4-23u^3+89u^2-142u+81}{u^2(u-2)^3\sqrt{\frac{2}{u}-1}}\cdot\arcsin\left(\sqrt{\frac{u}{2}}\right)\\
&=\frac{1}{30}u+\frac{16}{105}u^2+\frac{9}{35}u^3+\frac{1024}{3465}u^4+\frac{2500}{9009}u^5+\cdots\\
\end{align*}

$$ $$

Building blocks at $u=1$
We have now all the building blocks we need and derive some nice identities by setting $u=1$ and noting that $\arcsin\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}$.
\begin{align*}
A(1)&=\sum_{n=0}^\infty\frac{2^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}=1-\frac{1}{4}\pi\\
\left.(uD_u)A(u)\right|_{u=1}&=\sum_{n=0}^\infty\frac{n2^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}=-\frac{3}{2}+\frac{1}{2}\pi\\
\left.(uD_u)^2A(u)\right|_{u=1}&=\sum_{n=0}^\infty\frac{n^22^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}=\frac{5}{2}-\frac{3}{4}\pi\\
\left.(uD_u)^3A(u)\right|_{u=1}&=\sum_{n=0}^\infty\frac{n^32^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}=-\frac{7}{2}+\frac{5}{4}\pi\\
\left.(uD_u)^4A(u)\right|_{u=1}&=\sum_{n=0}^\infty\frac{n^42^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}=\frac{13}{2}-\frac{3}{2}\pi
\end{align*}

$$ $$

OPs series:
Since 
  \begin{align*}
&n[n(\pi^3+1)+\pi^2](n^2+n-1)\\
&\qquad=(1+\pi^3)n^4+(1+\pi^2+\pi^3)n^3-(1-\pi^2+\pi^3)n^2-\pi^2n
\end{align*}
we obtain by putting all together in OPs series
  \begin{align*}
\sum_{n=0}^{\infty}&{2^nn[n(\pi^3+1)+\pi^2](n^2+n-1)\over (2n+1)(2n+3){2n\choose n}}\\
&=\sum_{n=0}^{\infty}{2^n((1+\pi^3)n^4+(1+\pi^2+\pi^3)n^3-(1-\pi^2+\pi^3)n^2-\pi^2n)\over (2n+1)(2n+3){2n\choose n}}\\
&=2(1+\pi^3)\left.(uD_u)^4A(u)\right|_{u=1}+2(1+\pi^2+\pi^3)\left.(uD_u)^3A(u)\right|_{u=1}\\
&\qquad-2(1-\pi^2+\pi^3)\left.(uD_u)^2A(u)\right|_{u=1}-2\pi^2\left.(uD_u)A(u)\right|_{u=1}\\
&=2(1+\pi^3)\left(\frac{13}{2}-\frac{3}{2}\pi\right)+2(1+\pi^2+\pi^3)\left(-\frac{7}{2}+\frac{5}{4}\pi\right)
\\
&\qquad-2(1-\pi^2+\pi^3)\left(\frac{5}{2}-\frac{3}{4}\pi\right)-2\pi^2\left(-\frac{3}{2}+\frac{1}{2}\pi\right)\\
&=1+\pi+\pi^2+\pi^3+\pi^4
\end{align*}
  and the claim follows.