Find $\sum\limits_{t=0}^{\infty} \frac{1}{2^{n+2t}(n+2t)}\binom{n+2t}{t}$ First, we can obtain
\begin{align*} \frac{1}{2^{n+2t}(n+2t)}\binom{n+2t}{t}=\frac{(n+2t)!}{2^{n+2t}(n+2t)(n+t)!t!}&=\frac{(n+2t-1)!}{(2n+2t)!!(2t)!!}, \end{align*}
but this seems to be helpless. By the way, WA gives the result $\dfrac{1}{n}.$
 A: Using the integral representation for the inverse of the Beta function
\begin{equation}
 \int_{0}^{\pi/2}(\cos s)^{a-1}\cos\left(bs\right)\mathrm{d}s=\frac{\pi}{2^{a}}
\frac{1}{a\mathrm{B}\left(\frac{1}{2}(a+b+1),\frac{1}{2}(a-b+1)\right)}
\end{equation}
with $a=n+2t+1,b=n$, we have
\begin{align}
 \binom{n+2t}{t}&=\frac{1}{(n+2t+1)\mathrm{B}(n+t+1,t+1)}\\
 &=\frac{2^{n+2t+1}}{\pi}\int_{0}^{\pi/2}(\cos s)^{n+2t}\cos\left(ns\right)\mathrm{d}s
\end{align}
and thus
\begin{align}
 I&=\sum_{t=0}^{\infty} \frac{1}{2^{n+2t}(n+2t)}\binom{n+2t}{t}\\
&=\frac{2}{\pi}\sum_{t=0}^{\infty} \frac{1}{n+2t}\int_{0}^{\pi/2}(\cos s)^{n+2t}\cos\left(ns\right)\mathrm{d}s
\end{align}
For $n>0$, the integral can be evaluated by parts and the summation becomes a geometric series:
\begin{align}
 I&=\frac{2}{\pi n}\sum_{t=0}^{\infty}\int_{0}^{\pi/2}(\cos s)^{n+2t-1}\sin\left(ns\right)\sin s\,\mathrm{d}s\\
 &=\frac{2}{\pi n}\int_{0}^{\pi/2}\frac{\sin\left(ns\right)}{\sin s}(\cos s)^{n-1}\,\mathrm{d}s
\end{align}
Using the residue theorem, this integral can be shown to be independent of $n>0$ and equal to $\pi/2$. Finally,
\begin{equation}
 I=\frac{1}{n}
\end{equation}
as suggested.
A: From a CAS
$$\sum_{t=0}^{p} \frac{\binom{n+2t}{t}}{2^{n+2t}(n+2t)}=\frac{1}{n}-\frac{ \binom{n+2 p+2}{p+1} \,
   _3F_2\left(1,\frac{n+2p+2}{2},\frac{n+2p+3}{2};p+2,n+p+2;1\right)}{(n+2
   p+2)\, 2^{(n+2 p+2)}}$$ where appears a generalized hypergeometric function.
I am not able to find its asymptotics.
