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.
 A: We use an approach based upon the Lagrange inversion theorem and prove for $k$ being a non-negative integer
\begin{align*}
\color{blue}{\sum_{n=0}^\infty\binom{2n+k}{n}z^n=\left(\frac{2}{1+\sqrt{1-4z}}\right)^k\frac{1}{\sqrt{1-4z}}\qquad\qquad |z|<\frac{1}{4}}
\end{align*}
We follow thereby the paper Lagrange Inversion: when and how by R. Sprugnoli etal. We consider the generating function
\begin{align*}
F(z)=\sum_{n=0}^{\infty}a_nz^n=\sum_{n=0}^\infty\binom{2n+k}{n}z^n
\end{align*}
and apply formula (G6) from the paper. The formula (G6) tells us that if there are functions $A(u)$ and $\phi(u)$, so that the coefficient $a_n$ admits a representation
\begin{align*}
a_n=[u^n]A(u)\phi(u)^n
\end{align*}
where $[u^n]$ denotes the coefficient of $u^n$ of a series, then the following is valid:
\begin{align*}
F(z)&=\sum_{n=0}^{\infty}[u^n]A(u)\phi(u)^nz^n\\
&=\left.\frac{A(u)}{1-z\phi^{\prime}(u)}\right|_{u=z\phi(u)}\tag{1}
\end{align*}

We obtain
\begin{align*}
\color{blue}{F(z)}&\color{blue}{=\sum_{n=0}^\infty\binom{2n+k}{n}z^n}\\
&=\sum_{n=0}^\infty[u^n](1+u)^{2n+k}z^n\tag{2}\\
&=\left.\frac{(1+u)^k}{1-z\,\frac{d}{du}(1+u)^2}\right|_{u=z(1+u)^2}\tag{3}\\
&=\frac{(1+u)^k}{1-\frac{u}{(1+u)^2}\cdot 2(1+u)}\tag{4}\\
&=\frac{(1+u)^{k+1}}{1-u}\tag{5}\\
&=\frac{\left(\frac{2}{1+\sqrt{1-4z}}\right)^{k+1}}{2-\frac{2}{1+\sqrt{1-4z}}}\tag{6}\\
&\,\,\color{blue}{=\left(\frac{2}{1+\sqrt{1-4z}}\right)^k\frac{1}{\sqrt{1-4z}}}\tag{7}
\end{align*}
and the claim follows.

Comment:

*

*In (2) we use the coefficient of operator $[z^n]$ to denote the coefficient of a series. We observe we can apply the Lagrange Inversion theorem with $A(u)=(1+u)^k$ and $\phi(u)=(1+u)^2$.


*In (3) we use formula (1) with $A$ and $\phi$ as stated in the comment above.


*In (4) we substitute $z=\frac{u}{1+u}$ and do the derivation.


*In (5) we simplify the expression.


*In (6) we recall $u=z(1+u)^2$ which is a quadratic equation in $u=u(z)$. We calculate
\begin{align*}
u(z)&=\frac{1}{2z}\left(1-2z-\sqrt{1-4z}\right)\\
&=\frac{1}{2z}\left(1-\sqrt{1-4z}\right)-1\\
&=\frac{1}{2z}\,\frac{1-(1-4z)}{1+\sqrt{1-4z}}-1\\
&=\frac{2}{1+\sqrt{1-4z}}-1
\end{align*}
and take the solution with the minus sign, since this one can be expanded as generating function.


*In (7) we make again some simplifications to obtain the wanted form.

Note: The relationship with OPs first stated formula is given via
\begin{align*}
\color{blue}{\sum_{n=0}^\infty\binom{2n}{n+k}z^n}
&=\sum_{n=k}^\infty \binom{2n}{n+k}z^n\tag{8}\\
&=\sum_{n=k}^\infty \binom{2n}{n-k}z^n\tag{9}\\
&=\sum_{n=0}^\infty \binom{2n+2k}{n}z^{n+k}\tag{10}\\
&=z^k\left(\frac{2}{1+\sqrt{1-4z}}\right)^{2k}\frac{1}{\sqrt{1-4z}}\tag{11}\\
&\,\,\color{blue}{=\left(\frac{4z}{\left(1+\sqrt{1-4z}\right)^2}\right)^k\frac{1}{\sqrt{1-4z}}}
\end{align*}

Comment:

*

*In (8) we start the right-hand side with index $n=k$, since $\binom{2n}{n+k}=0$ if $n<k$.


*In (9) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.


*In (10) we shift the index to start with $n=0$ again.


*In (11) we use the identity (7) from above with $k\to 2k$ and make finally some simplifications.
A: 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.
