Infinite sum - generalising Catalan generating function

This is a follow up from this question where we proved that $$S = \sum_{n\geq 1}\frac{1}{n}\binom{2n}{n+1}2^{-2n} = 1$$

As a matter of facts this can be done using lots of (related) methodologies, including:

1. Maybe the easiest, using telescopic series by noting that $$\frac{1}{n}\binom{2n}{n+1}2^{-2n} = 2(b_n-b_{n+1}),\quad \text{with}\quad b_n=\binom{2n}{n}2^{-2n}$$
2. Using Gauss Hypergeometric theorem, from the fact that $$S+1 = {}_2F_1(\frac{1}{2},1,2,1)$$
3. Using Taylor / Binomial series, seeing that $$S= -2\left(\sum_{n\geq 0} \binom{1/2}{n}(-1)^{n}\right)+1$$
4. From Catalan number, seeing that this is the generating function for the Catalan numbers taken at value $$x=1/4$$, $$S = \frac{1}{2}c(1/4) = \frac{1}{2}\sum_{n=0}^\infty C_n(1/4)^n$$

I want to generalize the results further, showing that $$S_k = \sum_{n\geq k}\frac{k}{n}\binom{2n}{n+k}2^{-2n} = 1$$

I have tried to adapt the proofs for $$S$$, with no success so far. The fact that I'm no longer looking at the central binomial coefficient is always causing me trouble.

1. I can't find any telescopic series, I tried adapting $$b_n$$ or looking at some relation $$b_{n+k}-b_n$$.
2. I failed to reduce my sum to a Taylor series, or to find a hypergeometric function.
3. I've tried reasoning by induction. If I can express $$S_{k+1}$$ from $$S_k$$, this could work. But I end up with $$S_{k+1} = S_k + \sum_{n\geq k}2^{-2n}\binom{2n}{n+k}\frac{n-2k(k+1)}{(n+k+1)n}$$ And proving that this second sum is null seems harder than my original problem (it does go to 0 numerically, so this at least seems correct).
4. I also tried to bound $$S_{k+1}$$, and to prove that we must have $$S_{k+1}\geq S_k$$ (or the reverse), but my bounds are not tight enough.

I would appreciate any help or hint, maybe a direction to look at?

For instance I wonder if there are some numbers defined by difference of binomial coefficient, e.g. $$T_{n,k} = \binom{2n}{n+k}-\binom{2n}{n+k+1}$$ That would generalize Catalan's number (obtained with $$k=1$$ here). This might help my induction strategy.

$$S_k = \sum_{n\geq k}\frac{2k(2n-1)!}{(n-k)!(n+k)!}\cdot\frac{1}{4^n}=\sum_{n\geq k}\frac{(2n-1)!\left[(n+k)-(n-k)\right]}{(n-k)!(n+k)!}\cdot\frac{1}{4^n}$$ equals $$\sum_{n\geq k}\left[\frac{(2n-1)!}{(n-k)!(n+k-1)!}-\frac{(2n-1)!}{(n-1-k)!(n+k)!}\right]\frac{1}{4^n}$$ or $$\frac{1}{ 4^k}\sum_{m\geq 0}\left[\frac{(2m+2k-1)!}{m!(m+2k-1)!}-\frac{(2m+2k-1)!}{(m-1)!(m+2k)!}\right]\frac{1}{4^{m}}$$ or $$\frac{1}{4^k}\sum_{m\geq 0}\frac{1}{4^m}[x^m]\left[(1+x)^{2m+2k}\frac{1-x}{1+x}\right]$$ or $$\frac{1}{4^k}\sum_{m\geq 0}[x^m]\left[\left(\frac{1+x}{2}\right)^{2m}(1-x)(1+x)^{2k-1}\right]$$
or $$\frac{1}{4^k}\sum_{m\geq 0}\operatorname*{Res}_{x=0}\left[\left(\frac{1+x^2}{2x}\right)^{2m}\frac{(1-x^2)(1+x^2)^{2k-1}}{x}\right]$$ which finally boils down to something simple to compute (i.e. the opposite of the residue at infinity of a rational function), by switching $$\text{Res}$$ and $$\sum_{m\geq 0}$$. I am pretty sure I messed up some transformations but the approach in itself should be fine.
• Hi and thanks for the help, could you explain your notation $[x^m]$ and how you go from the third line to the fourth? Thanks – Thomas Lesgourgues Nov 1 '19 at 11:47
• @ThomasLesgourgues: $[x^n]f(x)$ is the coefficient of $x^n$ in the Maclaurin series of $f(x)$; the transition from the third to the fourth line is induced by $\binom{n}{k}=[x^k](1+x)^n$. – Jack D'Aurizio Nov 1 '19 at 17:08
The result is a direct consequence of this question : $$\sum_{n=0}^\infty\frac {(m+2n)!}{n!(m+n+1)!}(1/4)^n=2^{m+1}/(m+1) \qquad (1)$$
We have $$S_k = \sum_{n\geq k}\frac{k}{n}\binom{2n}{n+k}2^{-2n}$$ Therefore $$S_k = \sum_{n\geq 0}\frac{k}{n+k}2^{-2(n+k)}\frac{(2n+2k)!}{(n+2k)!n!}$$ $$S_k = k2^{-2k}\sum_{n\geq 0}2^{-2n}\frac{2\times(2n+2k-1)!}{(n+2k)!n!}$$ With $$m=2k-1$$ $$S_k = 2k2^{-2k}\sum_{n\geq 0}2^{-2n}\frac{(2n+m)!}{(n+m+1)!n!}$$ hence by $$(1)$$, $$S_k = 2k2^{-2k}\frac{2^{m+1}}{m+1}=1$$