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?
 A: 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.
A: 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.
A: Note that
\begin{align}
\sum_{n=1}^{+\infty}\left[e-\left(1+\frac{1}{n}\right)^{n}\right]\geq
&
\sum_{n=1}^{\infty}\left[\sum_{k=0}^{n}\frac{1}{k!}-\left(1+\frac{1}{n}\right)^{n}\right]
&
\quad \forall n\in\mathbb{N}
\\
=
&
\sum_{n=1}^{\infty}\left[\sum_{k=0}^{n}\frac{1}{k!}- \sum_{k=1}^{n}\frac{n!}{(n-k)!k!}\left(\frac{1}{n}\right)^k\right]
& \forall n\in\mathbb{N}
\\
=
&
\sum_{n=1}^{\infty}\sum_{k=0}^{n}\left[\frac{1}{k!}-\frac{n!}{(n-k)!k!}\left(\frac{1}{n}\right)^k\right]
& 
\forall n\in\mathbb{N}
\\
\geq
&
\sum_{n=1}^{\infty}
\left[\frac{1}{k!}-\frac{n!}{(n-k)!k!}\left(\frac{1}{n}\right)^k\right]
& 
\\
\geq
&
\sum_{n=1}^{\infty}
\left[\frac{1}{k!}\left(\frac{1}{n}\right)^k-\frac{n!}{(n-k)!k!}\left(\frac{1}{n}\right)^k\right]
& 
\\
\geq
&
\sum_{n=1}^{\infty}
\left[\frac{1}{k!}-\frac{n!}{(n-k)!k!}\right]\cdot\left(\frac{1}{n}\right)^k
& 
\end{align}
for all $k\in\{0,1,\dots,n\}$. If $k=1$, 
\begin{align}
\sum_{n=1}^{+\infty}\left[e-\left(1+\frac{1}{n}\right)^{n}\right]\geq
&
\sum_{n=1}^{\infty}
\left|1-n\right|\cdot\left(\frac{1}{n}\right)
&
\geq\sum_{n=1}^{\infty}\frac{1}{2}=\infty
\end{align}
