Answer
Divergent
Work Step by Step
To use the Direct Comparison Test, we need a known series $\Sigma_{n=1}^{\infty}b_{n}$ for the purpose of comparison.
We will choose $b_{n}=\frac{1}{n}$
The Comparison Test states that the p-series $\sum_{n=1}^{\infty}a_{n}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$.
Now, $a_{n}=\frac{1}{n^{1+\frac{1}{n}}}=\frac{1}{n.n^{\frac{1}{n}}}$ and $b_{n}=\frac{1}{n}$
$\lim\limits_{n \to \infty}\frac{a_{n}}{b_{n}}=\lim\limits_{n \to \infty}\frac{\frac{1}{n.n^{\frac{1}{n}}}}{\frac{1}{n}}$
$=\lim\limits_{n \to \infty}\frac{n}{n.n^{{1/n}}}$
$=\lim\limits_{n \to \infty}\frac{1}{n^{{1/n}}}$
$=\frac{1}{1}$ ( Since, $\lim\limits_{n \to \infty}n^{1/n}=1$ by L-Hospital's Rule)
$=1 \ne 0 \ne \infty$
The given series diverges by the Limit Comparison Test, because $\Sigma _{n=1}^{\infty}\frac{1}{n}$ is diverging.
