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

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.

Answer 1

Using Beta function, one can express:

$$\sum_{n=0}^{\infty}\frac{2^n(5n^5+5n^4+5n^3+5n^2-9n+9)}{(2n+1)(2n+2)(2n+3){2n\choose n}}$$
$$=\frac{5n^5+5n^4+5n^3+5n^2-9n+9}{(n+1)^2}\cdot\frac{1}{2}\int_0^1(2t(1-t))^{n+1}dt$$

Therefore the original expression can be transformed into:

$$\frac{1}{2}\int_0^1\sum_{n=0}^{\infty}\frac{(5n^5+5n^4+5n^3+5n^2-9n+9)(2t(1-t))^{n+1}}{(n+1)^2}dt$$

---

We know that
$$\sum_{n=0}^{\infty}\frac{t^n}{n+1}=-\frac{\ln(1-t)}{t}\tag1$$
Integrating, we get
$$\sum_{n=0}^{\infty}\frac{t^{n+1}}{(n+1)^2}=\text{Li}_2(t)\tag2$$
Differentiating, multiplying both sides by t and using the above formulas to do some algebra, we obtain
$$\sum_{n=0}^{\infty}n\frac{t^{n+1}}{(n+1)^2}=-\text{Li}_2(t)-\ln (1-t)\tag3$$
Repeating the process:
$$\sum_{n=0}^{\infty}n^2\frac{t^{n+1}}{(n+1)^2}=\frac{t \text{Li}_2(t)-\text{Li}_2(t)-t+2 t \ln (1-t)-2 \ln (1-t)}{t-1}\tag4$$
$$\sum_{n=0}^{\infty}n^3\frac{t^{n+1}}{(n+1)^2}=\frac{-t^2 \text{Li}_2(t)+2 t \text{Li}_2(t)-\text{Li}_2(t)+3 t^2-3 t^2 \ln (1-t)-2 t+6 t \ln (1-t)-3 \ln (1-t)}{(t-1)^2}\tag5$$
$$\sum_{n=0}^{\infty}n^4\frac{t^{n+1}}{(n+1)^2}=\frac{t^3 \text{Li}_2(t)-3 t^2 \text{Li}_2(t)+3 t \text{Li}_2(t)-\text{Li}_2(t)-6 t^3+4 t^3 \ln (1-t)+7 t^2-12 t^2 \ln (1-t)-3 t+12 t \ln (1-t)-4 \ln (1-t)}{(t-1)^3}\tag6$$
$$\sum_{n=0}^{\infty}n^5\frac{t^{n+1}}{(n+1)^2}=\frac{-t^4 \text{Li}_2(t)+4 t^3 \text{Li}_2(t)-6 t^2 \text{Li}_2(t)+4 t \text{Li}_2(t)-\text{Li}_2(t)+10 t^4-5 t^4 \ln (1-t)-14 t^3+20 t^3 \ln (1-t)+14 t^2-30 t^2 \ln (1-t)-4 t+20 t \ln (1-t)-5 \ln (1-t)}{(t-1)^4}\tag7$$

---

Set t $\mapsto$ $2t(1-t)$, substitute the above identities, and we simplify to get:

$=\frac{1}{2}\int_0^1\frac{18 \left(2 t^2-2 t+1\right)^4 \text{Li}_2(-2 (t-1) t)+\left(2 t^2-2 t+1\right)^4 \left(-\ln \left(2 t^2-2 t+1\right)\right)+20 \left(24 t^7-96 t^6+156 t^5-132 t^4+68 t^3-28 t^2+9 t-1\right) t}{\left(2 t^2-2 t+1\right)^4}dt$
$=\frac{1}{2}\underbrace{\int_0^118\text{Li}_2(-2 (t-1) t)dt}_{\large \color{blue}{\frac{9}{4}(-32+8\pi+\pi^2)}}-\frac{1}{2}\underbrace{\int_0^1\ln \left(2 t^2-2 t+1\right)dt}_{\large \color{red}{\frac{1}{2}(\pi-4)}}+\frac{1}{2}\underbrace{\int_0^1(30-\frac{90}{2t^2-2t+1}+\frac{130}{(2t^2-2t+1)^2}-\frac{100}{(2t^2-2t+1)^3}+\frac{30}{(2t^2-2t+1)^4})dt}_{\large \color{green}{-\frac{35}{2}(\pi-4)}}$
$=\frac{1}{2}(\color{blue}{\frac{9}{4}(-32+8\pi+\pi^2)}-\color{red}{\frac{1}{2}(\pi-4)}\color{green}{-\frac{35}{2}(\pi-4)})$
$=\LARGE \frac{9\pi^2}{8}$ $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$

---

Update: Someone asked me about the evaluation of $\int_0^1\text{Li}_2(-2 (t-1) t)dt$, and it can be done using integration by parts.

Evaluating $\int_0^1\text{Li}_2(-2 (t-1) t)dt$ \begin{align} & \int_0^1\text{Li}_2(-2t^2+2t)dt \\ & = \int_0^1\frac{-2t^2+t}{-t^2+t}\ln(1-2t+2t^2)dt \\ & = 2\int_0^1\ln(1-2t+2t^2)dt+\int_0^1\frac{-t}{-t^2+t}\ln(1-2t+2t^2)dt \\ & = -4+\pi-\frac{1}{2}\int_0^1\frac{\ln(1-2t+2t^2)}{-t^2+t}dt \\ & =-4+\pi-\frac{1}{2}(\int_0^1\frac{\ln(1-2t+2t^2)}{t}dt+\int_0^1\frac{\ln(1-2t+2t^2)}{1-t}dt) \\ & =-4+\pi-\int_0^1\frac{\ln(1-2t+2t^2)}{t}dt \\ & =\frac{1}{8}(-32+8\pi+\pi^2) \\ \end{align}

Answer 2

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.