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.
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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.
answered Jun 1, 2023 at 4:34
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4$\begingroup$ The problem with questions like this is that they often are raised at the same time at math.stackexchange, as it is the case here too. See math.stackexchange.com/questions/4710437. Answering such non-research questions (often in the imperative form "Show that ..." like here) encourages this misuse of MO. $\endgroup$ Commented Jun 1, 2023 at 10:38
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1$\begingroup$ @PeterMueller I agree, and sometimes I notice this before giving an answer. $\endgroup$ Commented Jun 1, 2023 at 15:52