Prove that:$\sum_{n=0}^{\infty}{2^{n+3}(n^2+n+\phi)\over (n+1)(2n+1)(2n+3){2n\choose n}}=\phi\pi^2+8\phi^2\pi-8\phi^3\sqrt{5}$

Question

$\phi$ is the golden ratio

$$\sum_{n=0}^{\infty}{2^{n+3}(n^2+n+\phi)\over (n+1)(2n+1)(2n+3){2n\choose n}}=\phi\pi^2+8\phi^2\pi-8\phi^3\sqrt{5}$$

I try:

$$S=\sum_{n=0}^{\infty}{2^{n+3}\over {2n\choose n}}\left[{A\over n+1}+{B\over 2n+1}+{C\over 2n+3}\right]$$

$n^2+n+\phi=A(2n+1)(2n+3)+B(n+1)(2n+3)+C(n+1)(2n+1)$

$A=-\phi$, $B=\phi-0.25$ and $C=\phi+0.75$

$$S=\sum_{n=0}^{\infty}{2^{n+3}\over {2n\choose n}}\left[{-\phi\over n+1}+{\phi-0.25\over 2n+1}+{\phi+0.75\over 2n+3}\right]$$

If we know the closed form

$$\sum_{n=0}^{\infty}{2^n\over {2n\choose n}(an+b)}=F(a,b)$$

Then it is easy to prove the above series.

I manage to find

$$\sum_{n=0}^{\infty}{2^n\over {2n\choose n}(2n+1)}={\pi\over 2}$$

Any help? Thank you.

Answer 1

Too long for a comment.

$$F(a,b)=\sum_{n=0}^{\infty}{2^n\over {2n\choose n}(an+b)}=\frac 1b \,\, _3F_2\left(1,1,\frac{b}{a};\frac{1}{2},\frac{b}{a}+1;\frac{1}{2}\right)$$ where appears the generalized hypergeometric function.

$$F(2,1)=\frac \pi 2$$ $$F(2,3)=\frac{3 \pi }{2}-4$$ $$F(1,1)=\pi -\frac{\pi ^2}{8}$$

Answer 2

Note that $$\frac{1}{\dbinom{2n}{n}}=\left(2n+1\right)B\left(n+1,n+1\right)=\left(2n+1\right)\int_{0}^{1}x^{n}\left(1-x\right)^{n}dx $$ hence $$\sum_{n\geq0}\frac{2^{n+3}}{\dbinom{2n}{n}\left(n+1\right)}=\sum_{n\geq0}\frac{2^{n+3}}{n+1}\left(2n+1\right)\int_{0}^{1}x^{n}\left(1-x\right)^{n}dx $$ $$=8\int_{0}^{1}\sum_{n\geq0}\frac{2n+1}{n+1}\left(2x\left(1-x\right)\right)^{n}=8\int_{0}^{1}\int_{0}^{1}\frac{1+2xy-2x^{2}y}{\left(1-2xy+2x^{2}y\right)^{2}}dxdy $$ $$=16\int_{0}^{1}\frac{dx}{1-2x\left(1-x\right)}+4\int_{0}^{1}\frac{\log\left(1-2x\left(1-x\right)\right)}{x\left(1-x\right)}dx $$ $$=\color{red}{8\pi-\pi^{2}}.$$ Note that the last integral can be calculated using the substitution $v=2x-1$. The other cases can be treated in the same way.