Looking for a proof of an interesting identity Working on a problem I have encountered an interesting identity:
$$
\sum_{k=0}^\infty \left(\frac{x}{2}\right)^{n+2k}\binom{n+2k}{k}
=\frac{1}{\sqrt{1-x^2}}\left(\frac{1-\sqrt{1-x^2}}{x}\right)^n,
$$
where $n$ is a non-negative integer number and $x$ is a real number with absolute value less than 1 (probably a similar expression is valid for arbitrary complex numbers $|z|<1$).
Is there any simple proof of this identity?

Extracting coefficients on the RHS we get the integral
(coefficient on $x^{n+2k}$)
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+2k+1}}
\frac{1}{\sqrt{1-z^2}}  
\left(\frac{1-\sqrt{1-z^2}}{z}\right)^n
\; dz.$$
Now we put $(1-\sqrt{1-z^2})/z = w$ so that $z = 2w/(1+w^2).$ This has
$w = \frac{1}{2} z + \cdots$ so the image in $w$ of the contour in $z$
can  be  deformed to  a  small  circle  enclosing  the origin  in  the
$w$-plane.  (Moreover  we see  that the  exponentiated term  starts at
$z^n$ which justifies the corresponding offset in the series.)  We get
$dz = 2/(1+w^2) - 4w^2/(1+w^2)^2 \; dw = 2(1-w^2)/(1+w^2)^2 \; dw.$ We
also have $1-z^2  = 1 - 4w^2/(1+w^2)^2 =  (1-w^2)^2/(1+w^2)^2.$ All of
this yields
$$\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{(1+w^2)^{n+2k+1}}{2^{n+2k+1} w^{n+2k+1}}
\frac{1}{(1-w^2)/(1+w^2)} w^n \frac{2(1-w^2)}{(1+w^2)^2} \; dw
\\ = \frac{1}{2^{n+2k}} \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{(1+w^2)^{n+2k}}{w^{2k+1}} \; dw.$$
This evaluates by inspection to
$$\frac{1}{2^{n+2k}} [w^{2k}] (1+w^2)^{n+2k}
= \frac{1}{2^{n+2k}} [w^{k}] (1+w)^{n+2k}
\\ = \frac{1}{2^{n+2k}} {n+2k\choose k}$$
which is the claim.
