Dtyjlui

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

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

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 $kge 0$

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 $(3)?$

What about reindexing and induction? The terms $frac{1}{(2n-1)cdots(2n-2k-1)}$ have a nice telescopic structure: by the residue theorem
$$ frac{1}{(2n-1)(2n-3)cdots(2n-2k-1)}=(-1)^ksum_{h=0}^{k}frac{(-1)^h}{(2n-2h-1)}cdotfrac{1}{2^{k+1}(2h)!!(2k-2h)!!} $$
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
$$ sum_{ngeq 1}frac{H_n}{4^{n}}binom{2n}{n}frac{1}{2n-2h-1}$$
through $frac{-log(1-z)}{1-z}=sum_{ngeq 1}H_n z^n$ and $frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}left(costhetaright)^{2n},dtheta$, 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 $dtheta$ and $dz$) and the Fourier series $sum_{mgeq 1}frac{cos(mvarphi)}{m}$ and $sum_{mgeq 1}frac{sin(mvarphi)}{m}$.

The only obstruction is that $frac{1}{2n-2h-1}=int_{0}^{1}z^nleft[frac{1}{2z^{h+3/2}}right],dz$ 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

$$ sum_{ngeq 1}frac{H_n binom{2n}{n}}{4^n(2n-1)} = sum_{ngeq 0}frac{2H_{n+1}binom{2n}{n}}{4^{n+1}(n+1)}=-frac{1}{pi}int_{0}^{1}sum_{ngeq 0}int_{0}^{pi/2}z^nleft(costhetaright)^{2n}log(1-z),dtheta,dz $$
or
$$ -frac{1}{pi}int_{0}^{1}int_{0}^{pi/2}frac{log(1-z)}{1-zcos^2theta},dtheta,dz =-frac{1}{2}int_{0}^{1}frac{log(1-z)}{sqrt{1-z}},dz,$$
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.