Answer
Converges to $e^{2}$
Work Step by Step
Given: $a_n=(1+\frac{2}{n})^{n}$
f(n)=$a_n$
let f(x)=$(1+\frac{2}{x})^{x}$
$\lim\limits_{n \to \infty}f(x)=\lim\limits_{x \to \infty}(1+\frac{2}{x})^{x}$
$\lim\limits_{n \to \infty}ln|f(x)|=\lim\limits_{x \to \infty}xln(1+\frac{2}{x})$
$\lim\limits_{n \to \infty}ln|f(x)|=\lim\limits_{x \to \infty}\frac{ln(1+\frac{2}{x})}{(\frac{1}{x})}$
Apply L'Hopital's rule
$\lim\limits_{n \to \infty}ln|f(x)|=\lim\limits_{x \to \infty}\frac{\frac{-2}{{x}^{2}}}{\frac{-1}{{x}^{2}}(1+\frac{2}{x})}$
$\lim\limits_{n \to \infty}ln|f(x)|=\lim\limits_{x \to \infty}\frac{2}{(1+\frac{2}{x})}$
$\lim\limits_{n \to \infty}ln|f(x)|=2$
Therefore $\lim\limits_{n \to \infty}f(x)={e}^{2}$
Therefore $\lim\limits_{n \to \infty}{(1+{\frac{2}{n})}^{n}}={e}^{2}$
