# How can we show that $\sum_{n=0}^{\infty}\frac{2^nn[n(\pi^3+1)+\pi^2](n^2+n-1)}{(2n+1)(2n+3){2n \choose n}}=1+\pi+\pi^2+\pi^3+\pi^4 ?$

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. $$\sum_{n=0}^{\infty}\frac{2^nn[n(\pi^3+1)+\pi^2](n^2+n-1)}{(2n+1)(2n+3){2n \choose n}}=1+\pi+\pi^2+\pi^3+\pi^4$$

• Have you tried expressing the terms of the sum as $f(n)-f(n+1)$? – Jack D'Aurizio Apr 30 '16 at 17:37

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 $y=\pi$

• somogyi can you help finished it. – user335850 May 1 '16 at 12:26
• Actually it can be finished from it using the same ideas, but the calculations required are quite tedious – Ákos Somogyi May 9 '16 at 18:50

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.

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:

For another solution approach using beta function, see my detailed solution at Prove $\sum_{n=0}^{\infty}{2^n(n^2-n\pi+1)(n^2+n-1)\over (2n+1)(2n+3){2n\choose n}}=1$ . The same steps can be followed with little changes in the coefficients.
Using Beta function, we can express $${2^n n[n(\pi^3+1)+\pi^2](n^2+n-1)\over (2n+1)(2n+3){2n\choose n}} = {n[n(\pi^3+1)+\pi^2](n^2+n-1)\over {n+1}}\int_{t=0}^1 T^{n+1} dt,$$ where $T=2t(1-t)$. Now, split these sums into forms $n^i T^{n+1}$ for $i=0,1,2,3$, and $T^{n+1}\over{n+1}$, whose sums are solved in the question link above. Then, integration w.r.t. $t$ will give the answer.