Answer
The series diverges by the nth term test.
Work Step by Step
$\Sigma_{n=1}^{\infty}(1+\frac{1}{n})^{n}$
Let's use the nth term test:
If $\lim\limits_{n \to \infty}a_n\ne0$, then the series diverges.
$\lim\limits_{n \to \infty}(1+\frac{1}{n})^{n}=1^\infty$
$1^\infty$ is an indeterminate form, but we cannot use L'Hopital's rule until the indeterminate form is either $\frac{0}{0}$ or $\frac{\infty}{\infty}$:
Use logarithmic differentiation:
$let$ $y=\lim\limits_{n \to \infty}(1+\frac{1}{n})^{n}$
$lny=\lim\limits_{n \to \infty}nln(1+\frac{1}{n})$
$lny=\lim\limits_{n \to \infty}\frac{ln(1+\frac{1}{n})}{\frac{1}{n}}=\frac{0}{0}$
$lny=\lim\limits_{n \to \infty}\frac{\frac{1}{1+n^{-1}}(-n^{-2})}{-n^{-2}}$
$lny=\lim\limits_{n \to \infty}\frac{1}{1+\frac{1}{n}}=1$
$e^{lny}=e^1$
$y=1$
Since $1\ne0$, the series diverges by the nth term test.
