Motivation:
This question will be inspired from:
Evaluation of $\sum\limits_{n=1}^\infty \frac 1{\text G(n)}≈ 3+\frac{\,_0\rm F_2(2,3;1)}2 $ with the Barnes G function? and Evaluating $\sum\limits_{x=2}^\infty \frac{1}{!x}$ in exact form.
with references from:
Evaluate $$\int_0^\infty\sum_{n=0}^\infty\frac{(-1)^n}{(n!)^t}\,dt$$
The Puzzle:
we will make heavy use of the Logarithm of the Gamma function which is technically its own function:
$$\text{ln}\Gamma(x)=\text{log}\Gamma(x)=\ln(\Gamma(x)$$
Here is the goal sum with $n\ge3$ or else the sum would diverge:
$$\sum_3^\infty \frac1{\text{ln}\Gamma(n)}$$
According to the reference question:
$$\sum_{n=2}^\infty\frac{(-1)^n}{\log\Gamma(n+1)}= \sum_{n=3}^\infty\frac{(-1)^{n-1}}{\log\Gamma(n)}= \int_0^\infty\sum_{n=0}^\infty\frac{(-1)^n}{(n!)^t}\,dt \implies -\int_0^\infty\sum_{n=0}^\infty\frac1{(n!)^t}dt\mathop = ^?\sum_3^\infty \frac1{\text{ln}\Gamma(n)} $$
Using this logic, let’s use the Hypergeometric function which may require $t\in\Bbb N$, but is still interesting:
$$-\int_0^\infty\sum_{n=0}^\infty\frac1{(n!)^t}dt=-\int_0^\infty\,_0\text F_{t-1}(1,…,1)dt$$
The first question is if the sum converges and the sum converges slowly. Maybe we can use a comparison test with a Stirling’s Approximation would work, but even it’s convergence is hard to use. Here are the partial sums:
Does our goal sum converge and if so, then what is an alternate form of it? Please correct me and give me feedback!
