Conjectured summation inequality

As @Mathphile pointed out in the comments, the series can be written as $$S=1+\sum_{n=1}^{\infty} \frac{(2n-1)!!}{(2n-2)!!\cdot(2n)^n} \left( \frac{1}{n} - \frac{1}{(n+1)^2} \right)=1+\sum_{n=1}^{\infty} \frac{a_nb_n}{2^{3n-2}}$$ where $$a_n=\frac{(2n-1)!}{n^n\cdot(n-1)!^2}\quad\text{and}\quad b_n=\frac{1}{n} - \frac{1}{(n+1)^2},$$ since $(2n-1)!!=(2n - 1)! / [2^{n - 1} \cdot (n - 1)!]$ and $(2n-2)!!=2^{n - 1} \cdot (n - 1)!$. Clearly the terms of $b_n$ form a decreasing sequence, and this also holds for $a_n$ since $$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{2\left(\frac2n+\frac1{n^2}\right)}{1+\frac1n}\left(1-\frac1{n+1}\right)^n<1.$$ Thus $$S=1+\sum_{n=1}^{10}\frac{a_nb_n}{2^{3n-2}}+\sum_{m>10}\frac{a_{10}b_{10}}{2^{3m-2}}<1+0.414+10^{-14}<\sqrt2.$$