Evaluating a generating function obtained from change of summation order I have an exercise involving snake oil to evaluate the sum $\sum_{k} {n \choose k} {2k \choose k} (-2)^{-k}$.  I wrote out the generating function $\sum_{n} \sum_{k} {n \choose k} {2k \choose k} (-2)^{-k} x^n$ and with some simplification I have come up with $\frac{1}{1-x} \sum_{k} {2k \choose k} \frac{x^k}{(2x-2)^k}$.  I am not sure how this sum simplifies as a generating function as I haven't worked with generating functions of this form.
 A: In general, we have $~\displaystyle\sum_{k\ge0}{2k\choose k}t^k=\frac1{\sqrt{1-4t}}~.~$ See binomial series. In our case, $t=\dfrac x{2(x-1)}$ .

$\displaystyle\sum_{k\ge0}{n\choose k}{2k\choose k}(-2)^{-k}~$ vanishes for odd values of n, and is equal to $~4^{-m}\displaystyle{2m\choose m},~$ for $n=2m$.
All in all, the final result is $~\dfrac1{\sqrt{1-x^2}}$ .
A: Since the range for $k$ is finite, and the binomials growing maximally polynomially, one can exchange the summation for $|x|<1$ as
\begin{align}
 \sum_{n}\sum_{k=0}^n\binom{2k}{k}\binom{n}{k}(-2)^{-k}x^n
&=\sum_{0\le k\le n}\binom{2k}{k}\binom{n}{k}(-2)^{-k}x^n\\
&=\sum_k\binom{2k}{k}(-2)^{-k}\sum_{n\ge k}\binom{n}{k}x^n\\
&=\sum_k\binom{2k}{k}(-2)^{-k}x^k(1-x)^{-k-1}\\
&=\frac1{1-x}\frac1{\sqrt{1-4\frac{x}{(-2)(1-x)}}}\\
&=\frac1{\sqrt{1-x^2}}
=\sum_m\binom{2m}{m}4^{-m}·x^{2m}
\end{align}
The combination of the square root series into the square root is valid for $2|x|/|1-x|<1$, i.e., $x\in(-1,\frac13)$. This is also compatible with the last series expansion, so that series identification is possible:
$$
\sum_{k=0}^n\binom{2k}{k}\binom{n}{k}(-2)^{-k}=
\begin{cases}
0&\text{ for odd }n\text{ and}\\
\binom{2m}{m}4^{-m}&\text{ for even }n=2m.
\end{cases}
$$

Used identities are
$$
\binom{n}{k}=(-1)^{n-k}\binom{-k-1}{n-k}\\
\binom{2k}{k}=(-4)^k\binom{-1/2}{k}
$$
that then allow to use Newton's binomial series.
