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?
 A: Using
 $$\binom{n}{k}=\frac{1}{2 \pi i}\oint_C\frac{(1+z)^{n}}{z^{k+1}}dz$$  we get (integration contour is the unit cicrle)
$$
2\pi iS_n=\oint dz \sum_{k=0}^{\infty}\frac{(1+z)^{n+2k}x^{n+2k}}{z^{k+1}2^{n+2k}}=\oint dz \frac{(1+z)^n x^n}{z2^n}\sum_{k=0}^{\infty}\frac{(1+z)^{2k}x^{2k}}{2^{2k}z^k}=\\
4\frac{x^n}{2^n}\oint dz \underbrace{\frac{(1+z)^n}{4z-(1+z)^2x^2}}_{f(z)}
$$
for $|x|<1$ only we have just one pole of $f(z)$ inside the unit circle namely $z_0(x)=\frac2{x^2}-\frac{2\sqrt{1-x^2}}{x^2}-1$ , so 
$$
S_n=4\frac{x^n}{2^n}\text{res}(f(z),z=z_0(x))=4\frac{x^n}{2^n}\left[ \frac{1}{4 \sqrt{1-x^2}}\left(2\frac{1-\sqrt{1-x^2}}{ x^2}\right)^n\right]
$$
or

$$
S_n=\frac{1}{\sqrt{1-x^2}}\left(\frac{1-\sqrt{1-x^2}}{ x}\right)^n
$$

A: 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.
A: I plan to insert this approach into the next (2019) version of my notes.
The even powers of $\arcsin(x)$ all have a nice Maclaurin series (see eqs (19),(20),(21),(22)), and this can be seen as a side effect of the Lagrange–Bürmann inversion theorem. I am going to implement the same trick here, by computing the derivatives at the origin of the RHS (the opposite way has already been investigated by tired).
$$[x^m]\left[\frac{1}{\sqrt{1-x^2}}\left(\frac{1-\sqrt{1-x^2}}{x}\right)^n\right]=\frac{1}{2\pi i}\oint\frac{(1-\sqrt{1-x^2})^n}{x^{n+m+1}\sqrt{1-x^2}}\,dx$$
where the integral is performed along a small circle enclosing the origin. What about enforcing the substitution $x=\sin z$? The sine function is holomorphic and injective in a neighbourhood of the origin, hence the RHS is simply converted into
$$ \frac{1}{2\pi i}\oint\frac{(1-\cos z)^n}{\left(\sin z\right)^{n+m+1}}\,dz =\frac{1}{2\pi i}\oint\frac{2^n \sin^{2n}\frac{z}{2}}{2^{n+m+1}\sin^{n+m+1}\frac{z}{2}\cos^{n+m+1}\frac{z}{2}}\,dz$$
or
$$\frac{1}{2\pi i}\oint\frac{\tan^{n-m-1}z}{2^{m}\cos^{2m}z}\,dz=\frac{1}{2^{m+1}\pi i}\oint u^{n-m-1}(1+u^2)^{m-1}\,du$$
via $z\to 2z$ and $z\to\arctan u$. Now the RHS is trivially given by a binomial coefficient multiplied by $\frac{1}{2^m}$ (provided by $m\geq n$) and the claim is proved.
