<s>[INST] Why does this infinite series equal one? $\sum_{k=1}^\infty \binom{2k}{k} \frac{1}{4^k(k+1)}=1$ Why does

$$\sum_{k=1}^\infty \binom{2k}{k} \frac{1}{4^k(k+1)}=1$$

Is there an intuitive method by which to derive this equality?
 [/INST] Your series is a telescoping one, since:
$$\begin{eqnarray*}\frac{1}{4^{k+1}}\binom{2k+2}{k+1}-\frac{1}{4^k}\binom{2k}{k}&=&\frac{1}{4^{k+1}}\binom{2k}{k}\left(\frac{(2k+2)(2k+1)}{(k+1)^2}-4\right)\\&=&-\frac{1}{2(k+1)4^{k}}\binom{2k}{k},\end{eqnarray*}$$
hence:
$$\sum_{k=1}^{+\infty}\binom{2k}{k}\frac{1}{4^k(k+1)}=2\sum_{k=1}^{+\infty}\left(\frac{1}{4^k}\binom{2k}{k}-\frac{1}{4^{k+1}}\binom{2k+2}{k+1}\right)=\frac{2}{4}\binom{2}{1}=1.$$
No need of generating functions or integrals.
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