General method of evaluating $\small\sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{n+k}$

Question

• Question: $ \mbox{How can we evaluate}\quad \sum_{n \geq 0}\left[{4^{n} \over \left(\, 2n + 1\,\right) \binom{2n}{n}}\right]^{2}{1 \over n + k}\quad \mbox{for general $k$ ?.} $

General methodology will be enough, but a closed-form is more preferable (if exists). Note that the previous problem, i.e. expressing binomial series in terms of MZVs, is solved via an alternative method (by user @pisco), so I simplified the question. For his method see here.

Answer 1

The problem is equivalent to finding an explicit form for a $\phantom{}_4 F_3$ with half-integer parameters, since due to Rodrigues' formula and Euler's Beta function

$$\small \int_{0}^{1}\!\!\!P_n(2x-1)\sum_{m\geq 0}\left(\frac{4^m}{(2m+1)\binom{2m}{m}}\right)^2 x^m\,dx=\!\!\int_{0}^{1}\!\!\sum_{m\geq n}\left(\frac{4^m}{(2m+1)\binom{2m}{m}}\right)^2\binom{m}{n}x^{m}(1-x)^n\,dx $$ equals $$ \frac{16^n}{(2n+1)^3\binom{2n}{n}^3}\cdot\phantom{}_4 F_3\left(n+1,n+1,n+1,n+1;n+\tfrac{3}{2},n+\tfrac{3}{2},2n+2;1\right).$$ Not surprising since $\phantom{}_4 F_3(1^{(4)};3/2^{(2)},2;x)$ essentially is the primitive of $\phantom{}_3 F_2(1^{(3)};3/2^{(2)};x)$.
Let us see if we manage to crack the case $n=0$: $$ \sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{n+1}=\sum_{n\geq 0}\frac{4^n}{(2n+1)(n+1)\binom{2n}{n}}\int_{0}^{\pi/2}\left(\sin\theta\right)^{2n+1}\,d\theta $$ due to the Maclaurin series of $\arcsin(x)^2$ equals $$ \int_{0}^{\pi/2}\frac{x^2}{\sin x}\,dx = 2\pi\, C-\frac{7}{2}\zeta(3)$$ and I guess this method can be applied to other values of $n$, too.
For instance, for $n=1$ we have to find $$ \sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{16(n+1)^3}{(n+2)(n+3)(2n+3)^2} $$ which by partial fractions decomposition boils down to evaluating $$\small\sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{n+A+1},\quad \sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{2n+2B+1},\quad \sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{(2n+2B+1)^2} $$ for specific values of $A,B\in\mathbb{N}$. The situation is the same for $n>1$.

---

A small collection of relevant identities: $$ \small\sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{n+2} = -\frac{1}{4}+\frac{\pi}{4}+\frac{\pi C}{2}-\frac{7\zeta(3)}{8} $$ $$ \small\sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{n+3} = -\frac{11}{64}+\frac{13\pi}{64}+\frac{9\pi C}{32}-\frac{63\zeta(3)}{128} $$ $$ \small\sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{2n+1} = -\pi\,C+\frac{7}{2}\zeta(3) $$ $$ \small\sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{2n+3} = -1+\frac{\pi}{2} $$ $$ \small\sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{1}{(2n+3)^2} = -3+\pi. $$

It is relevant to point out that $$ \frac{4^n}{(2n+1)\binom{2n}{n}}=\frac{2n+3}{2n+2}\cdot\frac{4^{n+1}}{(2n+3)\binom{2n+2}{n+1}} $$ so reindexing (together with the identities in this answer) is extremely useful for dealing with series of the last kind.

---

There is also this nice result that John Campbell, Marco Cantarini and I proved through fractional operators: if $f\in(C^{\omega}\cap L^2)(0,1)$ is such that $$ f(x)=\sum_{n\geq 0}a_n x^n = \sum_{m\geq 0} b_m P_m(2x-1) $$ then $$ \sum_{n\geq 0}\frac{a_n}{(2n+1)^2\left[\frac{1}{4^n}\binom{2n}{n}\right]^2} = \sum_{m\geq 0}\frac{(-1)^m b_m}{(2m+1)^2}.$$

Partial fraction decomposition then shows that for any $k\in\mathbb{Z}^+$ your series is easily converted into a linear combination with rational coefficients of $1,\pi,\pi C$ and $\zeta(3)$ through the FL-expansion of $\frac{1}{x^k}\left(-\log(1-x)-\sum_{s=1}^{k-1}\frac{x^k}{k}\right)$, which can be derived from $$ -\log(1-x)=1+\sum_{m\geq 1}(-1)^m\left(\frac{1}{m}+\frac{1}{m+1}\right)P_m(2x-1) $$ and the method previously outlined here.