How to prove the given series is divergent? Given series 
$$
\sum_{n=1}^{+\infty}\left[e-\left(1+\frac{1}{n}\right)^{n}\right],
$$ 
try to show that it is divergent!
The criterion will show that it is the case of limits $1$, so maybe need some other methods? any suggestions?

Let $x > 1$. Then the inequality $$\frac{1}{t} \leq \frac{1}{x}(1-t) + 1$$ holds for all $t \in [1,x]$ (the right hand side is a straight line between $(1,1)$ and $(x, \tfrac{1}{x})$ in $t$) and in particular $$\log(x) = \int_1^x \frac{dt}{t} \leq \frac{1}{2} \left(x - \frac{1}{x} \right)$$ for all $x > 1$.  Substitute $x \leftarrow 1 + \tfrac{1}{n}$ to get $$ \log \left(1 + \frac{1}{n} \right) \leq \frac{1}{2n} + \frac{1}{2(n+1)}$$ and after multiplying by $n$  $$\log\left(1 + \frac{1}{n} \right)^n \leq 1 - \frac{1}{2(n+1)}.$$ Use this together with the estimate $e^x \leq (1-x)^{-1}$ for all $x < 1$ to get $$\left(1 + \frac{1}{n} \right)^n \leq e \cdot e^{-\displaystyle\frac{1}{2(n+1)}} \leq e \cdot \left(1 - \frac{1}{2n+3} \right)$$  or $$e - \left(1 + \frac{1}{n} \right)^n \geq \frac{e}{2n+3}.$$  This shows that your series diverges.
