$\sum_{n=1}^{\infty} \left\{ e-(1+\frac{1}{n})^n \right\}$. is this converge or diverge

\begin{align*} &e - \left(1 + \frac1n\right)^n \\ = \, &\left(\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \ldots\right) - \left(\binom{n}{0} + \frac{1}{n}\binom{n}{1} + \frac{1}{n^2}\binom{n}{2} + \ldots + \frac{1}{n^n}\binom{n}{n}\right) \\ = \, &\frac{1 - \left(1 - \frac{1}{n}\right)}{2!} + \frac{1 - \left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)}{3!} + \ldots + \frac{1 - \left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) \ldots \left(1 - \frac{n - 1}{n}\right)}{n!} \\ + \, &\frac{1}{(n + 1)!} + \frac{1}{(n + 2)!} + \ldots \\ \ge \, &\frac{1 - \left(1 - \frac{1}{n}\right)}{2!} = \frac{1}{2n}, \end{align*} hence the series diverges.

Write $(1+ \frac{1}{n})^n=\exp(n \ln(1+\frac{1}{n}))$, develop as $n \to \infty$ and find an equivalent of $e-(1+ \frac{1}{n})^n$.