# 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 $$\pi<3+\frac{2}{9}$$.