check the convergence of the series

Question

So I have the following series: $$\sum_{n=2}^{\infty}\frac{e^n}{e^{n\sqrt[n]{n}}(ln(n))^2}$$ I thought that this series does not converge. $\sqrt[n]{n}\rightarrow 1$ so I thought that for large $n$ it will act like $\sum\frac{1}{ln(n)^2}$. But I checked this with wolframalpha and it told me that this series converges.

Where did I make a mistake?

Answer 1

Note that $\sqrt[n]{n}=\exp((\ln n)/n)\geqslant1+(\ln n)/n$ hence $n\sqrt[n]{n}\geqslant n+\ln n$ and the $n$th term of the series is at most $$\frac{\mathrm e^n}{\mathrm e^{n+\ln n}(\ln n)^2}=\frac1{n(\ln n)^2},$$ from which the convergence of the series follows.

Where did I make a mistake?

When you "deduced" from $\sqrt[n]{n}\to1$ (true) that $\mathrm e^{n\sqrt[n]{n}}/\mathrm e^n\to1$ (false).

Answer 2

Hint: Try Cauchy or Integral test.