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 ?$

Solution 1:

Generally the following sums may help: $$ \sum_{n=0}^\infty\frac{x^n}{\binom{2n}{n}}=4 \left(\frac{1}{4-x}+\frac{\sqrt{x} \arcsin\left(\frac{\sqrt{x}}{2}\right)}{(4-x)^{3/2}}\right) $$ $$ \sum_{n=0}^\infty\frac{nx^n}{\binom{2n}{n}}=\frac{\partial}{\partial x}\sum_{n=0}^\infty\frac{x^{n+1}}{\binom{2n}{n}}-\sum_{n=0}^\infty\frac{x^{n}}{\binom{2n}{n}} $$ $$ \sum_{n=0}^\infty\frac{x^n}{\binom{2n}{n}(2n+1)}=\frac{4 \arcsin \left(\frac{\sqrt{x}}{2}\right)}{\sqrt{(4-x) x}} $$ $$ \sum_{n=0}^\infty\frac{x^n}{\binom{2n}{n}(2n+3)}=-\frac{4 \left(2 \sqrt{(4-x) x}+(x-8) \sin ^{-1}\left(\frac{\sqrt{x}}{2}\right)\right)}{\sqrt{4-x} x^{3/2}}$$

Derivation of the third line from the first is as follows: $$ \sum_{n=0}^\infty\frac{x^n}{\binom{2n}{n}}=f(x) $$ $$ C+\sum_{n=0}^\infty\frac{x^{n}}{\binom{2n}{n}(2n+3)}=\frac{1}{x^3}\int\sum_{n=0}^\infty\frac{x^{n+2}}{\binom{2n}{n}}\mathrm{d}x=\frac{1}{x^3}\int f(x)x^2\mathrm{d}x $$ Where $C$ should be matched to the value at $x=0$+ there are some convergence criteria that should be checked. Also $$ \frac{n[n(y^3+1)+y^2](n^2+n-1)}{(2n+1)(2n+3)}=\frac{1}{2}\frac{n[n(y^3+1)+y^2](n^2+n-1)}{(2n+1)}-\frac{1}{2}\frac{n[n(y^3+1)+y^2](n^2+n-1)}{(2n+3)} $$ And then: $$ \frac{1}{2}\frac{n[n(y^3+1)+y^2](n^2+n-1)}{(2n+1)}=\frac{n^3 \left(8 y^3+8\right)+n^2 \left(-4 y^3+8 y^2-4\right)+n \left(-2 y^3-4 y^2-2\right)+3 y^3-2 y^2+3}{32}+\frac{-9y^3+6y^2-9}{2n+1} $$

Applying these, and a few more similarly obtained formulas, may help. There might be a better approach, with less calculation. (and there may be some mistakes). Finally plug in $x=2$ and $y=\pi$

Solution 2:

A quite boring approach. We can write your series as $$\left(\pi^{3}+1\right)\sum_{n\geq0}\frac{2^{n}n^{4}}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}}+\left(\pi^{3}+\pi^{2}+1\right)\sum_{n\geq0}\frac{2^{n}n^{3}}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}} $$ $$+\left(-\pi^{3}+\pi^{2}-1\right)\sum_{n\geq0}\frac{2^{n}n^{2}}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}}-\pi\sum_{n\geq0}\frac{2^{n}n}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}}. $$ Now using the identity $$\sum_{n\geq0}\frac{4^{n}x^{2n+2}}{\left(2n+1\right)\dbinom{2n}{n}}=\arcsin^{2}\left(x\right) $$ we have that, integrating both side and manipulating a bit, $$\sum_{n\geq0}\frac{2^{n}\left(\sqrt{2}x\right)^{2n}}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}}=\frac{2\sqrt{1-x^{2}}\arcsin\left(x\right)-2x+x\arcsin^{2}\left(x\right)}{x^{3}} $$ then if we differentiate $$-\pi\sum_{n\geq0}\frac{2^{n}n}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}}=-\frac{\pi}{2\sqrt{2}}\frac{d}{dx}\left.\frac{2\sqrt{1-x^{2}}\arcsin\left(x\right)-2x+x\arcsin^{2}\left(x\right)}{x^{3}}\right|_{x=1/\sqrt{2}} $$ $$ =-\frac{\pi}{2\sqrt{2}}\left.\frac{-6\sqrt{1-x^{2}}\arcsin\left(x\right)+6x-2x\arcsin^{2}\left(x\right)}{x^{4}}\right|_{x=1/\sqrt{2}}=\frac{\pi^{3}}{8}+\frac{3\pi^{2}}{2}-6\pi. $$ We can iterate this process and from $$\sum_{n\geq0}\frac{2^{n}n\left(\sqrt{2}x\right)^{2n}}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}}=\frac{-3\sqrt{1-x^{2}}\arcsin\left(x\right)+3x-x\arcsin^{2}\left(x\right)}{x^{3}} $$ we can differentiate again $$\sum_{n\geq0}\frac{2^{n}n^{2}}{\left(2n+1\right)\left(2n+3\right)\dbinom{2n}{n}}=\frac{1}{2\sqrt{2}}\frac{d}{dx}\left.\frac{-3\sqrt{1-x^{2}}\arcsin\left(x\right)+3x-x\arcsin^{2}\left(x\right)}{x^{3}}\right|_{x=1/\sqrt{2}} $$ and so on. I'm too lazy to complete the calculations so I leave the details to a willing person.

Solution 3:

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.

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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*}

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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*}

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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.