Show that the series
$$\sum\_{n=2}^{\infty}\frac{1}{[\frac{(1+\epsilon)\log n}{\log\log n}]!}$$converges for $\epsilon>0$.
Stirling's approximation gives that the convergence for the series is equivalent to the series
$$\sum\_{n=2}^{\infty}\frac{1}{\sqrt{2\pi [\frac{(1+\epsilon)\log n}{\log\log n}]}}\left(\frac{e}{[\frac{(1+\epsilon)\log n}{\log\log n}]}\right)^{[\frac{(1+\epsilon)\log n}{\log\log n} ]}.$$
But it is still hard to see the convergence.



convergence for a series
This is not a research level question, but I feel like answering it. Simply, use the elementary estimate
$$\log(m!)\sim m\log m.$$
The symbol $\sim$ means that the ratio of the two sides tends to $1$. For
$$m=\left[\frac{(1+\epsilon)\log n}{\log\log n}\right],$$
this becomes
$$\log(m!)\sim\frac{(1+\epsilon)\log n}{\log\log n}\log m\sim(1+\epsilon)\log n.$$
In particular, if $n$ is sufficiently large in terms of $\epsilon,$
then
$$m!>n^{1+\epsilon/2}.$$
Convergence is immediate from here.


