TITLE: Show that $\sum_{n=0}^{\infty}\frac{2^n(5n^5+5n^4+5n^3+5n^2-9n+9)}{(2n+1)(2n+2)(2n+3){2n\choose n}}=\frac{9\pi^2}{8}$
QUESTION [26 upvotes]: I don't how prove this series and I have try look through maths world and Wikipedia on sum for help but no use at all, so please help me to prove this series.
How to show that

$$\sum_{n=0}^{\infty}\frac{2^n(5n^5+5n^4+5n^3+5n^2-9n+9)}{(2n+1)(2n+2)(2n+3)\dbinom{2n}{n}}=\frac{9\pi^2}{8}$$

I have tried the stuff in the tags.

REPLY [10 votes]: Here is an answer based upon the arcsine function.

We start with the following formula valid for $u\in(0,2)$
  \begin{align*}
\sum_{n=0}^\infty&\frac{2^{n-1}}{(2n+1)(2n+3)\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.

$$ $$

We obtain by integrating this series (with some support of Wolfram Alpha) and a little rearrangement a series $A(u)$ which will serve as basis for further calculations.
  \begin{align*}
A(u)&=\sum_{n=0}^\infty\frac{2^{n}}{(2n+1)(2n+2)(2n+3)\binom{2n}{n}}u^n\\
&=-\frac{2}{u}+\frac{2}{u}\sqrt{\frac{2}{u}-1}\cdot\arcsin\left(\sqrt{\frac{u}{2}}\right)+\frac{1}{u}\left(\arcsin\left(\sqrt{\frac{u}{2}}\right)\right)^2\\
&=\frac{1}{6}+\frac{1}{60}u+\frac{1}{315}u^2+\frac{1}{1260}u^3+\frac{4}{17325}u^4+\frac{2}{27027}u^5+\cdots\\
\end{align*}

Since  we have to additionally respect a polynomial in $n$: $$5n^5+5n^4+5n^3+5n^2-9n+9$$ in the numerator of OPs series we apply the differential operator $D_u$ and calculate from $A(u)$
\begin{align*}
(uD_u)^kA(u)&=\sum_{n=0}^\infty\frac{n^k2^{n}}{(2n+1)(2n+2)(2n+3)\binom{2n}{n}}u^n\qquad\qquad k=1,\ldots,5
\end{align*}
These are  the building blocks to finally calculate OPs series.

Building blocks: $(uD_u)^kA(u)$
Successively applying the operator $uD_u$ on $A(u)$ we obtain (again with some help of Wolfram Alpha)
  \begin{align*}
(uD_u)A(u)&=\sum_{n=0}^\infty\frac{n2^{n}}{(2n+1)(2n+2)(2n+3)\binom{2n}{n}}u^n\\
&=\frac{3}{u}-\frac{3}{u}\sqrt{\frac{2}{u}-1}\cdot\arcsin\left(\sqrt{\frac{u}{2}}\right)
-\frac{1}{u}\left(\arcsin\left(\sqrt{\frac{u}{2}}\right)\right)^2\\
&=\frac{1}{60}u+\frac{2}{315}u^2+\frac{1}{420}u^3+\frac{16}{17325}u^4+\frac{10}{27027}u^5+\cdots\\
\\
(uD_u)^2A(u)&=\sum_{n=0}^\infty\frac{n^22^{n}}{(2n+1)(2n+2)(2n+3)\binom{2n}{n}}u^n\\
&=-\frac{9}{2u}-\frac{4-\frac{9}{u}}{u\sqrt{\frac{2}{u}-1}}\cdot\arcsin\left(\sqrt{\frac{u}{2}}\right)
+\frac{1}{u}\left(\arcsin\left(\sqrt{\frac{u}{2}}\right)\right)^2\\
&=\frac{1}{60}u+\frac{4}{315}u^2+\frac{1}{140}u^3+\frac{64}{17325}u^4+\frac{50}{27027}u^5+\cdots\\
\\
(uD_u)^3A(u)&=\sum_{n=0}^\infty\frac{n^32^{n}}{(2n+1)(2n+2)(2n+3)\binom{2n}{n}}u^n\\
&=\frac{13u-27}{2u(u-2)}+\frac{5u^2-24u+27}{u^2(u-2)\sqrt{\frac{2}{u}-1}}\cdot\arcsin\left(\sqrt{\frac{u}{2}}\right)
-\frac{1}{u}\left(\arcsin\left(\sqrt{\frac{u}{2}}\right)\right)^2\\
&=\frac{1}{60}u+\frac{8}{315}u^2+\frac{3}{140}u^3+\frac{256}{17325}u^4+\frac{250}{27027}u^5+\cdots\\
\\
(uD_u)^4A(u)&=\sum_{n=0}^\infty\frac{n^42^{n}}{(2n+1)(2n+2)(2n+3)\binom{2n}{n}}u^n\\
&=-\frac{3(6u^2-26u+27)}{2u(u-2)^2}\\
&\qquad-\frac{6u^3-47u^2+109u-81}{u^2(u-2)^2\sqrt{\frac{2}{u}-1}}\cdot\arcsin\left(\sqrt{\frac{u}{2}}\right)
+\frac{1}{u}\left(\arcsin\left(\sqrt{\frac{u}{2}}\right)\right)^2\\
&=\frac{1}{60}u+\frac{16}{315}u^2+\frac{9}{140}u^3+\frac{1024}{17325}u^4+\frac{1250}{27027}u^5+\cdots\\
\\
(uD_u)^5A(u)&=\sum_{n=0}^\infty\frac{n^52^{n}}{(2n+1)(2n+2)(2n+3)\binom{2n}{n}}u^n\\
&=\frac{24u^3-167u^2+352u-243}{2u(u-2)^3}\\
&\qquad-\frac{7u^4-82u^3+292u2-441u+243}{u^2(u-2)^3\sqrt{\frac{2}{u}-1}}\cdot\arcsin\left(\sqrt{\frac{u}{2}}\right)\\
&\qquad-\frac{1}{u}\left(\arcsin\left(\sqrt{\frac{u}{2}}\right)\right)^2\\
&=\frac{1}{60}u+\frac{32}{315}u^2+\frac{27}{140}u^3+\frac{4096}{17325}u^4+\frac{6250}{27027}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}}{(2n+1)(2n+2)(2n+3)\binom{2n}{n}}
=-2+\frac{1}{2}\pi+\frac{1}{16}\pi^2\\
\left.(uD_u)A(u)\right|_{u=1}&=\sum_{n=0}^\infty\frac{n2^{n}}{(2n+1)(2n+2)(2n+3)\binom{2n}{n}}
=3-\frac{3}{4}\pi-\frac{1}{16}\pi^2\\
\left.(uD_u)^2A(u)\right|_{u=1}&=\sum_{n=0}^\infty\frac{n^22^{n}}{(2n+1)(2n+2)(2n+3)\binom{2n}{n}}
=-\frac{9}{2}+\frac{5}{4}\pi+\frac{1}{16}\pi^2\\
\left.(uD_u)^3A(u)\right|_{u=1}&=\sum_{n=0}^\infty\frac{n^32^{n}}{(2n+1)(2n+2)(2n+3)\binom{2n}{n}}
=7-2\pi-\frac{1}{16}\pi^2\\
\left.(uD_u)^4A(u)\right|_{u=1}&=\sum_{n=0}^\infty\frac{n^42^{n}}{(2n+1)(2n+2)(2n+3)\binom{2n}{n}}
=-\frac{21}{2}+\frac{13}{4}\pi+\frac{1}{16}\pi^2\\
\left.(uD_u)^5A(u)\right|_{u=1}&=\sum_{n=0}^\infty\frac{n^52^{n}}{(2n+1)(2n+2)(2n+3)\binom{2n}{n}}
=17-\frac{19}{4}\pi-\frac{1}{16}\pi^2
\end{align*}

$$ $$

OPs series:
We obtain by putting all together in OPs series
  \begin{align*}
\sum_{n=0}^{\infty}&{2^n(5n^5+5n^4+5n^3+5n^2-9n+9)\over (2n+1)(2n+2)(2n+3){2n\choose n}}\\
&=5\left.(uD_u)^5A(u)\right|_{u=1}+5\left.(uD_u)^4A(u)\right|_{u=1}+5\left.(uD_u)^3A(u)\right|_{u=1}\\
&\qquad+5\left.(uD_u)^2A(u)\right|_{u=1}-9\left.(uD_u)A(u)\right|_{u=1}+9A(1)\\
&=5\left(17-\frac{19}{4}\pi-\frac{1}{16}\pi^2\right)+5\left(-\frac{21}{2}+\frac{13}{4}\pi+\frac{1}{16}\pi^2\right)\\
&\qquad+5\left(7-2\pi-\frac{1}{16}\pi^2\right)+5\left(-\frac{9}{2}+\frac{5}{4}\pi+\frac{1}{16}\pi^2\right)\\
&\qquad-9\left(3-\frac{3}{4}\pi-\frac{1}{16}\pi^2\right)+9\left(-2+\frac{1}{2}\pi+\frac{1}{16}\pi^2\right)\\
&=\frac{9}{8}\pi^2
\end{align*}
  and the claim follows.