Answer
The series is divergent.
Work Step by Step
$\sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{\infty}(1+\frac{1}{n})^{n^{2}}$
$|a_{n}|=(1+\frac{1}{n})^{n^{2}}$
$\lim\limits_{n \to \infty} \sqrt[n] {|a_{n}|}=\lim\limits_{n \to \infty} \sqrt[n] {(1+\frac{1}{n})^{n^{2}}}$
$=\lim\limits_{n \to \infty} {(1+\frac{1}{n})^{n}}$
$=\lim\limits_{n \to \infty} 1^{\infty}$
This is the form of $1^{\infty}$, so use L-Hospital's rule.
$=\lim\limits_{n \to \infty} e^{ln {(1+\frac{1}{n})^{n}}}$
$=e^{\lim\limits_{n \to \infty} nln {(1+\frac{1}{n})}}$
$=e\gt 1$
The series is divergent.