Answer
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^n}$ is convergent.
Work Step by Step
Let \[\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\frac{1}{n^n}\]
\[\Rightarrow a_n=\frac{1}{n^n}\]
Since, \[n^n\geq n^2\;\;\;\;\forall n\geq 1\]
\[\Rightarrow \frac{1}{n^n}\leq\frac{1}{n^2}\;\;\;\;\forall n\geq 1\]
[P-Series Test: $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^P}$ is convergent if and only if $P\geq 1$]
Also $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$ is convergent by P-Series test.
By Comparison Test, $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^n}$ is convergent.
