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

Question

I was looking at this paper on section [17],

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

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 $(3)?$

Answer 1

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 $k$.

On second thought, I have just applied the binomial transform in disguise.

Answer 2

Using Lemma $(13)$ in this paper provided by the OP $$\sum_{n=1}^{\infty}H_n{2n \choose n}x^n=\frac{2}{\sqrt{1-4x}}\ln\left(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}\right)$$ Relpace $x$ with $\displaystyle \frac{x^2}{4}$, we get $$\sum_{n=1}^{\infty}\frac{H_n}{4^n}{2n \choose n}x^{2n}=\frac{2}{\sqrt{1-x^2}}\ln\left(\frac{1+\sqrt{1-x^2}}{2\sqrt{1-x^2}}\right)$$ Now divide both sides by $x^2$ and integrate from $x=0$ to $x=1$, we get

\begin{align} S&=\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{4^n(2n-1)}=\int_0^1\frac{2}{x^2\sqrt{1-x^2}}\ln\left(\frac{1+\sqrt{1-x^2}}{2\sqrt{1-x^2}}\right)\ dx,\quad \color{blue}{x=\sin\theta}\\ &=2\int_0^{\pi/2}\csc^2\theta\ln\left(\frac{1+\cos\theta}{2\cos\theta}\right)\ d\theta\overset{IBP}{=}2\int_0^{\pi/2}\frac{d\theta}{1+\cos\theta}=2(1)=2 \end{align}