In continuation of this MO question: The sum of an hydrogen atom related infinite series.

Can the sum $$\sum\limits_{n=1}^{\infty}\frac{n-\frac{1}{2}}{n}\left[\frac{\Gamma(n)}{\Gamma\left(n+\frac{1}{2}\right)}\right]^4 \tag{1}$$ be calculated explicitly?

In the case of previous series, besides the explicit formula for the partial sums given by Robert Israel, the following simple reasoning works. Let $$a_n=\frac{1}{n+\frac{1}{2}}\left[\frac{\Gamma(n)}{\Gamma\left(n+\frac{1}{2}\right)}\right]^2.$$ Then $$\frac{a_n}{a_{n-1}}=\frac{(n-1)^2}{n^2-\frac{1}{4}}=\frac{4(n-1)^2}{4n^2-1}.$$ That is $$4n^2a_n=4(n-1)^2a_{n-1}+a_n.$$ Applying this recurrence relation again and again, we get $$4n^2a_n=4a_1+a_2+a_3+\ldots+a_n.$$ Therefore $$s_n=\sum\limits_{i=1}^na_i=4n^2a_n-3a_1,$$ and because $\lim\limits_{n\to\infty}n^2a_n=1$ and $a_1=\frac{8}{3\pi}$, we get $$s=\lim\limits_{n\to\infty}s_n=4-\frac{8}{\pi}.$$ However, in the case of (1), I was unable to elaborate the similar method.