Answer
Divergent
Work Step by Step
To use the Direct Comparison Test, we need to known series $\Sigma_{n=1}^{\infty}b_{n}$ for the purpose of comparison.
We will choose the harmonic series $b_{n}=\frac{1}{n}$
Now, $a_{n}=\frac{e^{1/n}}{n}$ and $b_{n}=\frac{1}{n}$
$\lim\limits_{n \to \infty}\frac{a_{n}}{b_{n}}=\lim\limits_{n \to \infty}\frac{\frac{e^{1/n}}{n}}{\frac{1}{n}}$
Since, $ e\approx 2.71 \gt 1$
$\sqrt[n] e=\sqrt[n] {2.71}\gt \sqrt[n] 1$
$\frac{e^{1/n}}{n} \gt \frac{1}{n}$
$=(1+0)^{2}\ne 1 \ne 0 \ne \infty $ for all $n$
The Comparison Test states that the p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$.
$\Sigma_{n=1}^{\infty}\frac{1}{n}$ is divergent by p- series Comparison Test with $p =1$ and so, $\Sigma \frac{e^{1/n}}{n}$ also diverges.
Hence, $\Sigma_{n=1}^{\infty} \frac{e^{1/n}}{n}$ diverges by the direct comparison test.
