What does this series converge to? $\sum_{n=0}^{\infty} (\frac{\gamma}{2})^{2n}\binom{2n}{\frac{k}{2}+n}$. Consider $\gamma \in (0,1)$ and $k \in 2\mathbb{N}$. Can we give a closed form expression for the series
$$\sum_{n=0}^{\infty} \left(\frac{\gamma}{2}\right)^{2n}\binom{2n}{\frac{k}{2}+n}?$$
Maybe if we choose a special value for $\gamma$, e.g. $\frac{1}{2}$?
(I obtained this series by summing up the transition probabilities of a random walk on the integers, starting at $0$, terminating at $k$ after $n$ steps, with $\gamma$ some discounting factor to ensure convergence everywhere. These probabilities, for even $k$, are given by $\frac{1}{2^n}\binom{n}{(k+n)/2}$ for $n$ even and $0$ for $n$ odd.)
 A: Since $k\in 2\Bbb N$ we have $m=k/2\in\Bbb N$. Now consider
$$
S(x,m)=\sum_{n=0}^{\infty}{2n \choose {m+n}} x^n.
$$
The binomial coefficient is equal to zero when $n<m$ so
$$
S(x,m)=\sum_{n=m}^{\infty}{2n \choose {m+n}} x^n=x^m\sum_{n=0}^{\infty}{2n+2m \choose {n+2m}} x^n
$$
Let $(s)_n=\Gamma(s+n)/\Gamma(s)$ be the Pochhammer symbol. We have by the gamma duplication formula and definition of the binomial coefficient:
$$
\begin{align}
{2n+2m \choose {n+2m}}
&=\frac{\Gamma(2n+2m+1)}{\Gamma(n+2m+1)n!}\\
&=\frac{2^{2n+2m}}{\sqrt\pi}\frac{\Gamma(m+1/2+n)\Gamma(m+1+n)}{\Gamma(n+2m+1)n!}\\
&=\frac{2^{2m}}{\sqrt\pi}\frac{\Gamma(m+1/2+n)\Gamma(m+1+n)}{\Gamma(2m+1+n)}\frac{4^n}{n!}\\
&=\underbrace{\frac{2^{2m}}{\sqrt\pi}\frac{\Gamma(m+1/2)\Gamma(m+1)}{\Gamma(2m+1)}}_{=1}%
\frac{\Gamma(m+1/2+n)\Gamma(m+1+n)\Gamma(2m+1)}{\Gamma(m+1/2)\Gamma(m+1)\Gamma(2m+1+n)}\frac{4^n}{n!}\\
&=\frac{(m+1/2)_n(m+1)_n}{(2m+1)_n}\frac{4^n}{n!}\\
\end{align}
$$
Substituting this result back into $S$ we have
$$
S(x,m)=x^m\sum_{n=0}^{\infty}\frac{(m+1/2)_n(m+1)_n}{(2m+1)_n}\frac{{4x}^n}{n!}.
$$
Finally, we substitute $x=\gamma^2/4$, $m=k/2$, and identify the series $S$ as a hypergeometric function to get
$$
S(\gamma^2/4,k/2)=\left(\frac{\gamma}{2}\right)^k{_2F_1}\left({\frac{k+1}{2},\frac{k}{2}+1\atop k+1};\gamma^2\right).
$$
According to this hypergeometric reduction formula, we then simplify our solution to get the final form
$$
\bbox[5px,border:2px solid #C0A000]{%
\sum_{n=0}^{\infty}{2n \choose {k/2+n}} \frac{\gamma^{2n}}{2^{2n}}
=\frac{\gamma^k}{\sqrt{1-\gamma^2}(1+\sqrt{1-\gamma^2})^k}.%
}
$$
