Show that $\sum_{n=0}^\infty \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} = 1.$

Famously, $\displaystyle \int_0^{\pi/2} \cos^{2n}{x}\,\mathrm{d}x = \frac{\pi}{2^{2n+1}}\binom{2n}{n}$ (e.g. see here); and $\displaystyle \frac{1}{1+n} = \int_0^1 y^n \, \mathrm{d} y$, thus:

$\displaystyle \begin{aligned} \sum_{n \ge 0}^\infty \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} & = \frac{1}{\pi} \sum_{n \ge 0}\int_0^{\pi/2}\int_0^1 y^n \cos^{2n}{x}\,\mathrm{d}y\,\mathrm{d}x\, \\& = \frac{1}{\pi} \int_0^{\pi/2}\int_0^1\sum_{n \ge 0} y^n \cos^{2n}{x}\,\mathrm{d}y\,\mathrm{d}x\, \\& = \frac{1}{\pi} \int_0^{\pi/2}\int_0^1\sum_{n \ge 0} (y\cos^2{x})^n\,\mathrm{d}y\,\mathrm{d}x\, \\& = \frac{1}{\pi}\int_0^{\pi/2}\int_0^1\frac{1}{1-y \cos^2{x}}\mathrm{d}y\,\mathrm{d}x \\& = \frac{1}{\pi} \int_0^{\pi/2}\sec^2{x} \cdot \log\left({\csc^2{x}}\right)\,\mathrm{d}x\, \\& = \frac{1}{\pi} \cdot \bigg[2x+\tan{x}\log(\sec^2{x})\bigg]_{x \to 0}^{x \to \pi/2} \\& = \frac{1}{\pi}\cdot \pi \\& = 1. \end{aligned} $

Probabilistically, the number $$2\,C_n\,\left(\frac{1}{2}\right)^{2(n+1)}=\frac{1}{n+1}\,\binom{2n}{n}\,\frac{1}{2^{2n+1}}$$ is the probability that a symmetric random walk on the lattice points of $\mathbb{R}$ will return to the starting point for the first time after $2(n+1)$ steps. However, it is not difficult to show that with probability $1$, this random walk returns to the starting point (see, for example, Theorem 3 of this link). This shows that $$\sum_{n=0}^\infty\,\frac{1}{n+1}\,\binom{2n}{n}\,\frac{1}{2^{2n+1}}=1.$$ The same idea can be used to verify the generating function of the Catalan numbers (by considering asymmetric random walks instead).

The Generalized Binomial Theorem says $$ \begin{align} (1-x)^{-1/2} &=1+\frac12\frac{x}{1!}+\frac12\!\cdot\!\frac32\frac{x^2}{2!}+\frac12\!\cdot\!\frac32\!\cdot\!\frac52\frac{x^2}{3!}+\cdots\\[6pt] &=\sum_{k=0}^\infty(2k-1)!!\frac{x^k}{2^kk!}\\ &=\sum_{k=0}^\infty\frac{(2k!)}{2^kk!}\frac{x^k}{2^kk!}\\ &=\sum_{k=0}^\infty\binom{2k}{k}\left(\frac x4\right)^k\tag1 \end{align} $$ Substituting $x\mapsto x/4$ gives $$ \frac1{\sqrt{1-4x}}=\sum_{k=0}^\infty\binom{2k}{k}x^k\tag2 $$ Integrating gives $$ \frac12-\frac12\sqrt{1-4x}=\sum_{k=0}^\infty\frac1{k+1}\binom{2k}{k}x^{k+1}\tag3 $$ Plug in $x=\frac14$ and multiply by $2$ $$ 1=\sum_{k=0}^\infty\frac1{k+1}\binom{2k}{k}\frac1{2^{2k+1}}\tag4 $$