# Sum of the series $\sum_{n=1}^\infty n^2e^{{-(\log n)}^{1+\delta}}$ for some $\delta > 0$.

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 $$\delta >0$$.

• $e^{2\log n-(\log n)^{1+\delta}} \le e^{-(\log n)^{1+\frac{\delta}{2}}} \le e^{-2\log n} = n^{-2}$ for large $n$ – mathworker21 Nov 15 '18 at 6:13
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 $$\alpha>1$$.