How to find the limit $\lim_{n\to \infty}\left(n-\frac{n}{e}\left(1+\frac{1}{n}\right)^n\right)$ To find the limit
$$\displaystyle\lim_{n\to \infty}\left(n-\frac{n}{e}\left(1+\frac{1}{n}\right)^n\right)$$
 A: Hint: Using the Taylor series of the logarithm and the exponential function
\begin{align*}
\left( {1 + \frac{1}{n}} \right)^n & = \exp \left( {n\log \left( {1 + \frac{1}{n}} \right)} \right)  = \exp \left( {1 - \frac{1}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right) \\ & = e\exp \left( { - \frac{1}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right) = e\left( {1 - \frac{1}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right).
\end{align*}
A: Let $x=1/n$, then
$$L=\lim_{x\to 0}\left(\frac{1}{x}-\frac{1}{ex}(1+x)^{1/x}\right)$$
Use the McLaurin series: $$(1+x)^{1/x}=e-ex/2+11ex^2/24+...$$
$$L=\lim_{x \to 0} \left( \frac{1}{x}-\frac{1}{x}+\frac{1}{2}-\frac{11x}{24}+...\right)=\frac{1}{2}.$$
A: To get even more than the limit itself, compose Taylor series one piece at the time $$y=n-\frac{n}{e}\left(1+\frac{1}{n}\right)^n$$
$$a=\left(1+\frac{1}{n}\right)^n \implies \log(a)=n \log\left(1+\frac{1}{n}\right)$$
$$\log(a)=n\left(\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{3 n^3}+O\left(\frac{1}{n^4}\right)\right)$$
$$a=e^{\log(a)}=e-\frac{e}{2 n}+\frac{11 e}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ making
$$y=\frac{1}{2}-\frac{11}{24 n}+O\left(\frac{1}{n^2}\right)$$
