<s>[INST] Find a closed-form solution to the following summation I am solving a summation that appears in a paper, it claims that
$$\sum_{n=1}^{\infty} \binom{2n}{n+k}z^n=\bigg(\frac{4z}{(1+\sqrt{1-4z})^2}\bigg)^k$$
I found this identity here in equation (66)
$$\sum_{n=0}^{\infty} \binom{2n+k}{n}z^n=1+\sum_{n=1}^{\infty} \binom{2n+k}{n}z^n=1+\sum_{n=1}^{\infty} \binom{2n+k}{n+k}z^n =\bigg(\frac{2}{1+\sqrt{1-4z}}\bigg)^k\frac{1}{\sqrt{1-4z}}$$
I didn't found the paper that has the proof of the identity above, but I notice that
$$\binom{2n}{n+k}+\sum_{j=0}^{k-1} \binom{2n+j}{n+k-1}=\binom{2n+k}{n+k}$$
Therefore we obtain
$$\sum_{n=1}^{\infty} \binom{2n}{n+k}z^n=\bigg(\frac{2}{1+\sqrt{1-4z}}\bigg)^k\frac{1}{\sqrt{1-4z}}-1-\sum_{n=1}^{\infty} \sum_{j=0}^{k-1} \binom{2n+j}{n+k-1}z^n$$
Yet I have no idea for the last summation above, any help would be appreciated.
 [/INST] The identity
$$\sum_{n\ge 0} {2n+k\choose n} z^n
= \left(\frac{2}{1+\sqrt{1-4z}} \right)^k
\frac{1}{\sqrt{1-4z}}
= Q_k(z)$$
can also be proved with the Cauchy Coefficient Formula, radius of
convergence is the  distance to the nearest singularity which is at
$1/4.$
We obtain
$$[z^n] Q_k(z) =
\frac{1}{2\pi i}
\int_{|z|=\varepsilon}
\frac{1}{z^{n+1}}
\left(\frac{2}{1+\sqrt{1-4z}} \right)^k
\frac{1}{\sqrt{1-4z}} \; dz.$$
Now we put $w=\sqrt{1-4z}$ with the branch cut on $[1/4,\infty)$
(principal branch of the logarithm) so that we have
analyticity in a neighborhood of the origin and $dw = -2
\frac{1}{\sqrt{1-4z}} \; dz$. With $\sqrt{1-4z} = 1-2z-\cdots$ the image
of $|z|=\varepsilon$ makes  one turn around $w=1$ plus lower order
fluctuations so that it may  be deformed to a small circle and we get
with $z=(1-w^2)/4$
$$- \frac{1}{2} \frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{4^{n+1}}{(1-w^2)^{n+1}}
\left(\frac{2}{1+w} \right)^k \; dw
\\ = - \frac{2^{2n+k+1}}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(1-w)^{n+1}} \frac{1}{(1+w)^{n+1+k}} \; dw
\\ = \frac{(-1)^n\times 2^{n}}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(w-1)^{n+1}} \frac{1}{(1+(w-1)/2)^{n+1+k}} \; dw$$
This yields by the Cauchy Residue Theorem
$$(-1)^n 2^n \times (-1)^n \frac{1}{2^n} {n+n+k\choose n}
= {2n+k\choose n}$$
as claimed.
Remark. This identity and the proof appeared at the following MSE link from which it may be obtained by a straightforward square root manipulation.
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