Note, This is a re-post. My last question was poorly worded and I do not think it will be seen well.
How can you show the convergence of $\displaystyle{\int_{1}^{\infty} {\mathrm{d}x \over \left(x!\right)^{\,\epsilon}}\,,\ \forall\ \epsilon > 0\,}$ if it's at all possible.
My thoughts and attempt was to first replace $x!$ with $\Gamma(x+1)$, since $x!$ only maps naturals to naturals. So, $$ \int_{1}^{\infty}{\mathrm{d}x \over \Gamma\left(x + 1\right)^{\,\epsilon}} $$ And now attempt to do the comparrison test, say, $2^{x}$. I could not find a function that is everywhere less than the factorial function that also converges, maybe i'm missing something. But here is my attempt.
I found the first integer in the domain $[1,\infty)$ such that $2^{x} < \Gamma(x+1)$ which I got to be $x=4$. Then I took the integral and split it up.
$$\int_{1}^{\infty} \frac {1} {{\Gamma(x+1)}^{\epsilon}}dx = \int_{1}^{4} \frac {1} {{\Gamma(x+1)}^{\epsilon}}dx + \int_{4}^{\infty} \frac {1} {{\Gamma(x+1)}^{\epsilon}}dx$$
Now, knowing that $2^x$ is less than $\Gamma(x+1)$ on the interval $[4,\infty)$, we now have that $$\frac {1} {{\Gamma(x+1)}^{\epsilon}} < \frac {1}{2^{x\epsilon}}$$ and so $$\int_{4}^{\infty} \frac {1} {{\Gamma(x+1)}^{\epsilon}}dx < \int_{4}^{\infty} \frac {1} {{2}^{x\epsilon}}dx < \infty$$
And now finally, because $\int_{1}^{4} \frac {1} {{\Gamma(x+1)}^{\epsilon}}dx$ certainly converges to some finite value, and $\int_{4}^{\infty} \frac {1} {{\Gamma(x+1)}^{\epsilon}}dx$ converges to some other finite value by the comparison, then adding them together should show that the entire integral $$\int_{1}^{\infty} \frac {1} {{\Gamma(x+1)}^{\epsilon}}dx < \infty$$ and converges for any $\epsilon >0$ and then in a sense, so must $\frac{1}{x!}$
Correct?
$\Gamma(x+1)$ is a log-convex function by the Bohr-Mollerup theorem. It follows that:
$$ \forall x\geq 1,\qquad \log\Gamma(x+1)\geq \log\Gamma(1+1)+(x-1)\log\Gamma'(1+1)=(1-\gamma)(x-1) $$ and by exponentiating back, for any $c>0$: $$ \Gamma(x+1) \geq e^{(1-\gamma)(x-1)},\qquad \int_{1}^{+\infty}\frac{dx}{\Gamma(x+1)^c} \leq \frac{1}{c(1-\gamma)}. $$