Infinity limit

$\sqrt[n]{e^{n-1}}=e^{\frac{n-1}n}<n$ for $n>1$ . Because $e^{\frac{n-1}n}$ has upper bound of $e$ so for $n\geq3$ , it's trivial and we can manually check that it is true for $n=2$ .

So we have $\dfrac{1}{\sqrt[n]{e^{n-1}}}=\dfrac{1}{e^{\frac{n-1}n}}>\dfrac1n$ , hence $\displaystyle \sum_{n=2}^N \dfrac{1}{e^{\frac{n-1}n}} > \sum_{n=2}^N\dfrac1n$ $$ If $\displaystyle \sum_{n=2}^\infty\dfrac1n$ diverges, then $\displaystyle \frac{1}{\sqrt{e} } +\frac{1}{\sqrt[3]{e^{2}} }+\frac{1}{\sqrt[4]{e^{3}} } +\cdots = \sum_{n=2}^\infty\dfrac{1}{e^{\frac{n-1}n}}$ must also diverges.