How to calculate $\sum_{n=0}^{\infty}C_n\left(\frac{1}{2}\right)^{2n+1}\left(\frac{n+1}{2n+1}\right)$ Does the following sum equal 1 (or some amount less than 1)?
$$S\equiv\sum_{n=0}^{\infty}C_n\left(\frac{1}{2}\right)^{\!2n+1}\!\!\left(\frac{n+1}{2n+1}\right)=\sum_{n=0}^{\infty}\left(\frac{(2n)!}{(n+1)!\cdot n!}\right)\left(\frac{1}{2}\right)^{\!2n+1}\!\!\left(\frac{n+1}{2n+1}\right)$$
where $C_n$ is the $n$th Catalan number.
The first 100 sums yields .7573;
The first 1000 sums yields .7765;
The first 10000 sums yields .7826;
The first 100000 sums yields .7845. It is not clear to me if $S=1$ or $S<1$.
I know the following:
$$\sum_{n=0}^{\infty}C_n\left(\frac{1}{2}\right)^{\!2n+1}=1$$

Simplifying a bit the notation, since $C_n=\frac{1}{n+1}\binom{2n}{n}$, we want to compute
$$ S = \frac{1}{2}\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\int_{0}^{1}x^{2n}\,dx \tag{1}$$
and we may easily recognize the Maclaurin series of $\frac{1}{\sqrt{1-x^2}}$, leading to:
$$ S = \frac{1}{2}\int_{0}^{1}\frac{dx}{\sqrt{1-x^2}}=\frac{\arcsin 1}{2}=\color{red}{\frac{\pi}{4}}.\tag{2}$$
Notice that $\frac{1}{4^n}\binom{2n}{n}\leq\frac{1}{\sqrt{\pi n}}$ implies, through creative telescoping,
$$ S\leq \frac{1}{2}\left(1+\frac{1}{\sqrt{\pi}}\sum_{n\geq 1}\frac{1}{(2n+1)\sqrt{n}}\right)\stackrel{\text{CT}}{\leq}\frac{1}{2}\left(1+\frac{1}{\sqrt{\pi}}\sum_{n\geq 1}\frac{1}{\sqrt{n-\frac{1}{6}}}-\frac{1}{\sqrt{n+\frac{5}{6}}}\right)$$
or $S\leq \frac{1}{2}\left(1+\sqrt{\frac{6}{5\pi}}\right)$, from which it follows that $\pi<3+\frac{2}{9}$.
