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.
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 $\alpha>1$.
