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Is the series $$\sum_{n=1}^\infty n^2e^{{-(\log n)}^{1+\delta}}$$ convergent for some $\delta > 0$?

I tried to do it by comparison test but it was not doable. I tried to take help from Infinite series $\sum _{n=2}^{\infty } \frac{1}{n \log (n)}$. I was unable to proceed further. What I have got is that $$\lim_{n\to \infty}n^2e^{{-(\log n)}^{1+\delta}} = 0$$ for any $\delta >0$.

Thanks in advance for your suggestions.

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  • $\begingroup$ $e^{2\log n-(\log n)^{1+\delta}} \le e^{-(\log n)^{1+\frac{\delta}{2}}} \le e^{-2\log n} = n^{-2}$ for large $n$ $\endgroup$ – mathworker21 Nov 15 '18 at 6:13
  • $\begingroup$ Wow. Thank you @mathworker21. $\endgroup$ – TRUSKI Nov 15 '18 at 6:24
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Yes, it does converge for some $\delta>0$. Here is an example. I assume that by $\log$ you mean natural logarithm.
Note that $e^{-\log(n)^{1+\delta}}=e^{-(\log \,n)(\log\,n)^\delta}=n^{-(\log \,n)^\delta}$. Then your series is $\sum_{n=1}^\infty n^{2-(\log \,n)^\delta}$. Take $\delta=1$. Since $(\log\,20)-2<1$ and $(\log\,21)-2>1$ we have $$ \sum_{n=1}^\infty n^{2-\log \,n} = \sum_{n=1}^\infty {1\over n^{(\log \,n)-2}} = \sum_{n=1}^{21} {1\over n^{(\log \,n)-2}} + \sum_{n=22}^\infty {1\over n^{(\log \,n)-2}} $$ $$ \le \sum_{n=1}^{21} {1\over n^{(\log \,n)-2}} + \sum_{n=22}^\infty {1\over n^{(\log \,22)-2}} < \infty. $$ The last series converges because $\sum_{n\ge 1} n^{-\alpha}$ converges if $\alpha>1$.

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