Answer
Converges
Work Step by Step
Determine whether the series converges or diverges.
$\Sigma$$\frac{1}{n^{n}}$
Consider $\frac{1}{n^{n}}$ $\leq$ $\frac{1}{n^{2}}$ and by P - series for $\frac{1}{n^{2}}$ p > 1 and converges.
Therefore by comparison $\frac{1}{n^{n}}$ also converges.