Looking at this sum $\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\frac{{2n \choose n}}{4^n}g(n)$

Question

I am just observing this sum type $$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\frac{{2n \choose n}}{4^n}g(n)\tag1$$

I tried varies functions of $g(n)$ then I came upon this one

$g(n)=\frac{2n+1}{(2n+1)^4-(2k)^4}$ which does give a neat closed form, but I can't prove it.

$$\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\cdot\frac{{2n \choose n}}{4^n}\cdot\frac{2n+1}{(2n+1)^4-(2k)^4}=-\frac{\pi\tanh(k\pi)}{64k^3}\tag2$$

where $k\ge 1$

How do we prove $(2)?$

Answer 1

Since $$ \begin{align} \frac{(2n)!!}{(2n+1)!!} &=\frac{2^nn!}{\frac{(2n+1)!}{2^nn!}}\\ &=\frac1{2n+1}\frac{4^n}{\binom{2n}{n}} \end{align} $$ we get $$ \frac{(2n)!!}{(2n+1)!!}\frac{\binom{2n}{n}}{4^n}=\frac1{2n+1} $$ Therefore, using Partial Fractions and $(7)$ from this answer, $$ \begin{align} \sum_{n=0}^\infty\frac1{(2n+1)^4-x^4} &=\frac1{2x^2}\sum_{n=0}^\infty\left[\frac1{(2n+1)^2-x^2}-\frac1{(2n+1)^2+x^2}\right]\\ &=\frac1{4x^3}\sum_{n=0}^\infty\left[\frac1{(2n+1)-x}-\frac1{(2n+1)+x}\right]\\ &-\frac1{4ix^3}\sum_{n=0}^\infty\left[\frac1{(2n+1)-ix}-\frac1{(2n+1)+ix}\right]\\ &=-\frac1{8x^3}\sum_{n\in\mathbb{Z}}\frac1{n+\frac{1+x}2}+\frac1{8ix^3}\sum_{n\in\mathbb{Z}}\frac1{n+\frac{1+ix}2}\\ &=-\frac\pi{8x^3}\cot\left(\pi\frac{1+x}2\right)+\frac\pi{8ix^3}\cot\left(\pi\frac{1+ix}2\right)\\[6pt] &=\frac\pi{8x^3}\tan\left(\frac{\pi x}2\right)-\frac\pi{8x^3}\tanh\left(\frac{\pi x}2\right) \end{align} $$ Plug in $x=2k$.

Answer 2

Your summand simplifies to $\frac{1}{(2n+1)^4-(2k)^4}$, and from there you have $\sum_{n=0}^{\infty}\frac{1}{(2n+1)^4-(2k)^4}=\frac{\pi[\tan(\pi k)-\tanh(\pi k)]}{64k^3}$. You get your result from assuming $k\in\mathbb{Z}$.