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?

Using the binomial theorem, we get that for $n\ge1$,
$$
\begin{align}
\left(1+\frac1n\right)^{\large n}
&=\sum_{k=0}^n\frac{\binom{n}{k}}{n^k}\\
&=\sum_{k=0}^\infty\frac1{k!}\frac{n}{n}\frac{n-1}{n}\frac{n-2}{n}\cdots\frac{n-k+1}{n}\\
&\le2+\left(1-\frac1n\right)\sum_{k=2}^\infty\frac1{k!}\\
&=2+\left(1-\frac1n\right)(e-2)\\
&=e-\frac{e-2}{n}
\end{align}
$$
Therefore, for $n\ge1$,
$$
e-\left(1+\frac1n\right)^{\large n}\ge\frac{e-2}{n}
$$
So by comparison with the Harmonic series,
$$
\sum_{n=1}^\infty\left(e-\left(1+\frac1n\right)^{\large n}\right)
$$
diverges.
