Vrftjkry

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 $$lim_{nto infty}n^2e^{{-(log n)}^{1+delta}} = 0$$ 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
$$
sum_{n=1}^infty n^{2-log ,n}
=
sum_{n=1}^infty {1over n^{(log ,n)-2}}
=
sum_{n=1}^{21} {1over n^{(log ,n)-2}}
+
sum_{n=22}^infty {1over n^{(log ,n)-2}}
$$
$$
le
sum_{n=1}^{21} {1over n^{(log ,n)-2}}
+
sum_{n=22}^infty {1over n^{(log ,22)-2}} < infty.
$$
The last series converges because $sum_{nge 1} n^{-alpha}$ converges if $alpha>1$.