TITLE: Summation of series $\sum_{k=0}^\infty 2^k/\binom{2k+1}{k}$
QUESTION [9 upvotes]: 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?

REPLY [3 votes]: 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