Yesterday I got (but I haven't tested it numerically) that
$$\sum_{n=1}^\infty \sum_{k=1}^\infty\frac{G_n}{k}\frac{\Gamma(k+\frac{3}{2})\Gamma(n+\frac{1}{2})}{(n+k+1)!}=\frac{\pi(1-8\log 2)}{8},$$
where $G_n$ denotes the sequence of Gregory coefficients or reciprocal logarithmic numbers, see the Wikipedia [*Gregory coefficients*](https://en.wikipedia.org/wiki/Gregory_coefficients). Using the generating function for this sequence and the result of the Problem 4383 from Crux Mathematicorum ([1]), I can then write the closed-form of a series of the type 


$$\sum_{n=1}^\infty \sum_{k=1}^\infty(\text{a function of }G_n\text{, or }|G_n|,\text{ and }k)\cdot(\text{particular values of the gamma function}).$$

>**Question.** Are known series of previous type and how are evaluated? If it is in the literature refer it, and I try to search and read those statements from the literature. In other case, what work can be done to deduce similar formulas to get the closed-form for double series of the previous type? **Many thanks.**

Here I add my example of series, how I got the result for the series of the first paragraph.

**Example.** All issues of convergence are satisfied by uniform convergence of the series in the unit interval. Multiplying the Maclaurin series of the Gregory coefficients by $\sqrt{\frac{x}{1-x}}\log x$ and integrating the equation over the unit interval, one gets invoking the Problem 4383, due to Michel Bataille, that $$\sum_{n=1}^\infty G_n\left(\int_0^1(\log x)\sqrt{\frac{x}{1-x}}x^n dx\right)=\frac{\pi(-1+8\log 2)}{8}.\tag{1}$$
We change the varialbe of the integral $x=1-z$ and use the Taylor series of the logarithm to state that $$\sum_{n=1}^\infty \sum_{k=1}^\infty\frac{G_n}{k}\left(\int_0^1 z^k (1-z)^n \sqrt{\frac{1-z}{z}}dz\right)=\frac{\pi(1-8\log 2)}{8},\tag{2}$$
and finally I've used Wolfram Alpha online calculator (it is the definition of the gamma function) to get a closed-form of the integral in terms of particular values of the gamma function.

References:
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[1] *Problem 4383*, Crux Mathematicorum, Volume 45, Number 6, July 2019.