## Calculus: Early Transcendentals 8th Edition

$\sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{\infty}(1+\frac{1}{n})^{n^{2}}$ $|a_{n}|=(1+\frac{1}{n})^{n^{2}}$ $\lim\limits_{n \to \infty} \sqrt[n] {|a_{n}|}=\lim\limits_{n \to \infty} \sqrt[n] {(1+\frac{1}{n})^{n^{2}}}$ $=\lim\limits_{n \to \infty} {(1+\frac{1}{n})^{n}}$ $=\lim\limits_{n \to \infty} 1^{\infty}$ 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.