Generalizing $\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)}=2$

I was looking at this paper on section ,

$$\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)}=2\tag1$$

Let generalize $$(1)$$

$$\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)\cdots [2n-(2k+1)]}\tag2$$

Where $$k\ge 0$$

I conjectured the closed form for $$(2)$$ to be

$$\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)(2n-3)\cdots [2n-(2k+1)]}=\frac{2(-1)^k}{(2k+1)!!(2k+1)}\tag3$$

Here are a first few values of $$k=1,2$$ and $$3$$

\begin{align} \sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)(2n-3)}&=-\frac{2}{9}\tag4\\ \sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)}&=\frac{2}{75}\tag5\\ \sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)(2n-7)}&=-\frac{2}{735}\tag6 \end{align}

How do we go about to prove this conjecture $$(3)?$$

• How did you determine (4), (5) and (6)? Have you tested them numerically? – Richard Jan 2 at 18:21
• I tested them numerically. it seems correct – Endgame Jan 2 at 18:30

What about reindexing and induction? The terms $$\frac{1}{(2n-1)\cdots(2n-2k-1)}$$ have a nice telescopic structure: by the residue theorem $$\frac{1}{(2n-1)(2n-3)\cdots(2n-2k-1)}=(-1)^k\sum_{h=0}^{k}\frac{(-1)^h}{(2n-2h-1)}\cdot\frac{1}{2^{k+1}(2h)!!(2k-2h)!!}$$ equals $$\frac{(-1)^k}{2^{2k+1}k!} \sum_{h=0}^{k}\frac{(-1)^h}{(2n-2h-1)}\binom{k}{h}.$$ The natural temptation is now to compute $$\sum_{n\geq 1}\frac{H_n}{4^{n}}\binom{2n}{n}\frac{1}{2n-2h-1}$$ through $$\frac{-\log(1-z)}{1-z}=\sum_{n\geq 1}H_n z^n$$ and $$\frac{1}{4^n}\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{\pi/2}\left(\cos\theta\right)^{2n}\,d\theta$$, multiply both sides by $$(-1)^k \binom{k}{h}$$, sum over $$h=0,1,\ldots,k$$ and finish by invoking Fubini's theorem (allowing to switch the integrals with respect to $$d\theta$$ and $$dz$$) and the Fourier series $$\sum_{m\geq 1}\frac{\cos(m\varphi)}{m}$$ and $$\sum_{m\geq 1}\frac{\sin(m\varphi)}{m}$$.
The only obstruction is that $$\frac{1}{2n-2h-1}=\int_{0}^{1}z^n\left[\frac{1}{2z^{h+3/2}}\right]\,dz$$ does not hold unconditionally: we would have been happier in having rising Pochhammer symbols rather than falling ones. On the other hand, reindexing fixes this issue. Since $$\binom{2n+2}{n+1} = \frac{2(2n+1)}{n+1}\binom{2n}{n}$$, the original series can be written as
$$\sum_{n\geq 1}\frac{H_n \binom{2n}{n}}{4^n(2n-1)} = \sum_{n\geq 0}\frac{2H_{n+1}\binom{2n}{n}}{4^{n+1}(n+1)}=-\frac{1}{\pi}\int_{0}^{1}\sum_{n\geq 0}\int_{0}^{\pi/2}z^n\left(\cos\theta\right)^{2n}\log(1-z)\,d\theta\,dz$$ or $$-\frac{1}{\pi}\int_{0}^{1}\int_{0}^{\pi/2}\frac{\log(1-z)}{1-z\cos^2\theta}\,d\theta\,dz =-\frac{1}{2}\int_{0}^{1}\frac{\log(1-z)}{\sqrt{1-z}}\,dz,$$ clearly given by a derivative of the Beta function. This approach works also by replacing $$(2n-1)$$ with $$(2n-1)\cdots(2n-2k-1)$$, you just have to be careful in managing the involved constants depending on $$k$$.