Summation of series $\sum_{k=0}^\infty 2^k/\binom{2k+1}{k}$ How to find the sum of this series? 
$$\sum_{k=0}^{\infty}\cfrac{{2}^{k}}{\binom{2k+1}{k}}$$
It seems very easy. But I still can not work it out, can anyone help? 
 A: We have, using the Euler Beta function:
$$\begin{eqnarray*}\color{red}{\sum_{k=0}^{+\infty}\frac{2^k}{\binom{2k+1}{k}}}&=&\sum_{k=0}^{+\infty}\frac{2^k \Gamma(k+1)\Gamma(k+2)}{\Gamma(2k+2)}=\sum_{k=0}^{+\infty}2^k(k+1)\,B(k+1,k+1)\\&=&\sum_{k=0}^{+\infty}2^k(k+1)\int_{0}^{1}x^k(1-x)^k\,dx\\&=&\int_{0}^{1}\frac{dx}{(1-2x(1-x))^2}=\color{red}{\frac{\pi}{2}+1}.\end{eqnarray*}$$
A: This question is closely related to this question.
$$
\begin{align}
\sum_{k=0}^\infty\frac{2^k}{\binom{2k+1}{k}}
&=\sum_{k=0}^\infty2^k\frac{k!(k+1)!}{(2k+1)!}\tag{1}\\
&=\sum_{k=0}^\infty\frac{(k+1)!}{(2k+1)!!}\tag{2}\\
&=\sum_{k=0}^\infty\frac{k!}{(2k+1)!!}
+\sum_{k=0}^\infty\frac{k\,k!}{(2k+1)!!}\tag{3}\\
&=\sum_{k=0}^\infty\frac{k!}{(2k+1)!!}
+\sum_{k=0}^\infty\left(\frac{k!}{(2k-1)!!}-\frac{(k+1)!}{(2k+1)!!}\right)\tag{4}\\
&=\frac\pi2+1\tag{5}
\end{align}
$$
Explanation:
$(1)$: rewrite binomial coefficient with factorials
$(2)$: $(2k+1)!!=\frac{(2k+1)!}{2^kk!}$
$(3)$: $(k+1)!=k!+k\,k!$
$(4)$: $\frac{k!}{(2k-1)!!}-\frac{(k+1)!}{(2k+1)!!}=\frac{(2k+1)k!}{(2k+1)!!}-\frac{(k+1)k!}{(2k+1)!!}=\frac{k\,k!}{(2k+1)!!}$
$(5)$: this answer and telescoping series
