Find the value $\sum_{n=1}^{\infty}(e-(1+\frac{1}{n})^n)$ How to find the following series' value?

$$\sum_{n=1}^{\infty}\bigg(e-\Big(1+\frac{1}{n}\Big)^n\bigg)$$ 


Taylor expansion of $\frac {e^{x} - 1} {x}$ around $x = 0$ is $$1 + \frac {x} {2} + \frac {x^2} {3} + \cdots > 1 + \frac {x} {2}.$$ Put $n = 1/x$ and let $n \to \infty$, we have $$n \left(e^{1/n} - \left(1 + \frac {1} {n}\right)\right) > \frac {1} {2n}.$$ By Minkowski inequality, we have $$e - \left(1 + \frac {1} {n}\right)^n \geqslant n \left(e^{1/n} - \left(1 + \frac {1} {n}\right)\right) > \frac {1} {2n}.$$ But $\sum \frac {1} {2n}$ diverges, then so does your series.
