Limit with criteria $\lim_{n \to \infty}n \cdot \left [ \frac1e \left (1+\frac{1}{n+1} \right )^{n+1}-1 \right ]$ $$\lim_{n \to \infty}n \cdot \left [ \frac{\left (1+\frac{1}{n+1} \right )^{n+1}}{e}-1 \right ]$$
I was trying to calculate a limit that drove me to this case of Raabe-Duhamel's test, but I don't know how to finish it. Please give me a hint or a piece of advise.
I cannot use any of the solution below, but they are clear and good. I'm trying to prove it using squeeze theorem like this:
$$\lim_{n \to \infty}n \cdot \left [ \frac{\left (1+\frac{1}{n+1} \right )^{n+1}}{e}-1 \right ]=\frac{-1}{e} \cdot\lim_{n \to \infty}n \cdot \left [e- \left (1+\frac{1}{n+1} \right )^{n+1} \right ]$$ 
I found this:
$$\frac{e}{2n+2}<e- \left (1+\frac{1}{n} \right )^{n}<\frac{e}{2n+1}$$
Is this true? How can I prove this? Thanks for the answers.
 A: HINT
As suggested in the comments by Lord Shark the Unknown, we can use that
$$\left (1+\frac{1}{n+1} \right )^{n+1}=e^{(n+1)\log\left (1+\frac{1}{n+1} \right )}=e^{(n+1)\left (\frac{1}{n+1}-\frac{1}{2(n+1)^2}+o\left(\frac1{n^2}\right) \right )}$$
A: An overkilled method:
\begin{align*}
n\left[\dfrac{\left(1+\dfrac{1}{n+1}\right)^{n+1}}{e}-1\right]&=\dfrac{n}{n+1}\cdot(n+1)\cdot\left[\dfrac{\left(1+\dfrac{1}{n+1}\right)^{n+1}}{e}-1\right],
\end{align*}
so we are to look at 
\begin{align*}
(n+1)\cdot\left[\dfrac{\left(1+\dfrac{1}{n+1}\right)^{n+1}}{e}-1\right],
\end{align*}
somehow it is the same looking at
\begin{align*}
n\cdot\left[\dfrac{\left(1+\dfrac{1}{n}\right)^{n}}{e}-1\right]=\dfrac{1}{e}\cdot n\cdot\left[\left(1+\dfrac{1}{n}\right)^{n}-e\right].
\end{align*}
We note that
\begin{align*}
\lim_{x\rightarrow 0}(1+x)^{1/x}=e,
\end{align*}
so
\begin{align*}
n\cdot\left[\left(1+\dfrac{1}{n}\right)^{n}-e\right]=n\int_{0}^{1/n}\left((1+x)^{1/x}\right)'dx.
\end{align*}
Taking the derivative of the integrand, we find that
\begin{align*}
n\int_{0}^{1/n}\left((1+x)^{1/x}\right)'dx=n\int_{0}^{1/n}\left((1+x)^{1/x}\left(-\dfrac{1}{x^{2}}\log(1+x)+\dfrac{1}{x}\dfrac{1}{x+1}\right)\right)dx.
\end{align*}
By Integral Mean Value Theorem applied to the interval $[0,1/n]$, we get
\begin{align*}
&n\int_{0}^{1/n}\left((1+x)^{1/x}\left(-\dfrac{1}{x^{2}}\log(1+x)+\dfrac{1}{x}\dfrac{1}{x+1}\right)\right)dx\\
&=(1+\eta_{x})^{1/\eta_{x}}\left(-\dfrac{1}{\eta_{x}^{2}}\log(1+\eta_{x})+\dfrac{1}{\eta_{x}}\dfrac{1}{\eta_{x}+1}\right).
\end{align*}
However, it is not hard to compute that 
\begin{align*}
\lim_{x\rightarrow 0}(1+x)^{1/x}\left(-\dfrac{1}{x^{2}}\log(1+x)+\dfrac{1}{x}\dfrac{1}{x+1}\right)=-e/2,
\end{align*}
so as $x\rightarrow 0$, $\eta_{x}\rightarrow 0$ and the limit is $-1/2$.
A: hint
Since $$\lim_{n\to\infty}\frac{n}{n-1}=1,$$
It is the same to compute
$$\lim_{n\to\infty}n(\frac{(1+\frac 1n)^{n}}{e}-1)$$
use the fact that 
$$n\ln(1+\frac 1n)=n(\frac 1n -\frac{1}{2n^2} +\frac{1}{n^2}\epsilon(n))$$
$$=1-\frac{1}{2n}+\frac 1n\epsilon(n).$$
