How to calculate $\sum_{n=1}^{\infty}A_n(1/2)^{2n+2}(n+2)/(2n+2)$ This is a follow-up to a previous problem, which was answered as follows:
$$S=\sum_{n=0}^{\infty}C_n\left(\frac{1}{2}\right)^{\!2n+1}\!\!\left(\frac{n+1}{2n+1}\right)=\frac{\pi}{4},$$
where $C_n$ is the $n$th Catalan number.
A related problem has arisen which asks for the following summation:
$$T=\frac{1}{2}+\sum_{n=1}^{\infty}A_n\left(\frac{1}{2}\right)^{\!2n+2}\!\!\left(\frac{n+2}{2n+2}\right),$$
where $A_n$ is integer sequence A000245 (a "Catalan sequence convolved with its shifted variant") given by:
$$A_n=3\frac{(2n)!}{(n+2)!\cdot(n-1)!}.$$
Any thoughts on how to proceed? I am wondering if $S<T$.

Assuming that $A_n$ is defined as $\frac{3(2n)!}{(n+\color{red}{2})!(n-1)!}$ (otherwise the associated series is divergent, since $\frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi n}}$) we have that $T$ equals
$$ \frac{1}{2}+\frac{3}{8}\sum_{n\geq 1}\frac{1}{4^n}\binom{2n}{n}\frac{n}{(n+1)^2} $$
which due to $\sum_{n\geq 0}\frac{x^n}{4^n}\binom{2n}{n}=\frac{1}{\sqrt{1-x}}$ can be written as
$$ T = \frac{1}{2}+\frac{3}{8}\int_{0}^{1}\left(\frac{1}{\sqrt{1-x}}-1\right)(1+\log x)\,dx = \color{red}{-\frac{1}{4}+\frac{3}{2}\log 2}.$$
