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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?

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We have, using the Euler Beta function:

$$\begin{eqnarray*}\color{red}{\sum_{k=0}^{+\infty}\frac{2^k}{\binom{2k+1}{k}}}&=&\sum_{k=0}^{+\infty}\frac{2^k \Gamma(k+1)\Gamma(k+2)}{\Gamma(2k+2)}=\sum_{k=0}^{+\infty}2^k(k+1)\,B(k+1,k+1)\\&=&\sum_{k=0}^{+\infty}2^k(k+1)\int_{0}^{1}x^k(1-x)^k\,dx\\&=&\int_{0}^{1}\frac{dx}{(1-2x(1-x))^2}=\color{red}{\frac{\pi}{2}+1}.\end{eqnarray*}$$

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

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