What is the limit of $\lim_{n\to +\infty} n\left(\frac{e}{\left(1+\frac{1}{n}\right)^n}-1\right)$

Question

I have read this post Solve the following limit: $\lim_{n->\infty} n(\frac{\frac{1}{n!}(\frac{n}{e})^n}{\frac{1}{(n+1)!}(\frac{n+1}{e})^{n+1}}-1)$ but I don't understand how do you get from $\log\left(1+\tfrac{1}{n}\right)^n = 1-\frac{1+o(1)}{2n}$ to $\left(1+\tfrac{1}{n}\right)^n = e -\frac{e+o(1)}{2n}$ and then to $\frac{e}{\left(1+\frac{1}{n}\right)^n} = 1+\frac{1+o(1)}{2n}$ from there is obviously simple but maybe there is a way to solve the limit in an easier fashion.

Answer 1

We are using that as $x \to 0$

$$\log (1+x)=x-\frac12 x^2+o(x^2)$$

and

$$e^{1+x}=e\cdot e^{x}=e\cdot (1+x+o(x))$$

and

$$\frac1{1+x}=1-x+o(x)$$

Answer 2

If you are ok with L'Hospital you can calculate the limit relatively straight forward. It isn't really easier but also an option.

You may use the substitution $n= \frac{1}{t}$ and consider the limit for $t\to 0^+$ as follows:

\begin{eqnarray*} n\left(\frac{e}{\left(1+\frac{1}{n}\right)^n}-1\right) & = & \underbrace{\frac{1}{\left(1+\frac{1}{n}\right)^n}}_{\stackrel{n\to \infty}{\longrightarrow}\frac{1}{e}}\cdot \frac{e - \left(1+\frac{1}{n}\right)^n}{\frac{1}{n}}\\ & \stackrel{n=\frac{1}{t}}{\sim} &\frac{1}{e}\frac{e - (1+t)^{\frac{1}{t}}}{t}\\ & \stackrel{L'Hosp.}{\sim} &\frac{1}{e}\underbrace{\frac{(1+t)^{\frac{1}{t}}}{1+t}}_{\stackrel{t\to 0^+}{\longrightarrow}e}\frac{(1+t)\ln (1+t) - t}{t^2} \\ & \stackrel{L'Hosp.}{\sim} &\frac{\ln (1+t)}{2t} \\ & \stackrel{L'Hosp.}{\sim} &\frac{1}{2(1+t)} \stackrel{t\to 0^+}{\longrightarrow} \frac{1}{2} \end{eqnarray*}

Answer 3

Let $x=1/n$, then $$L=\lim_{x \rightarrow 0} \frac{1}{x} \left(\frac{e}{(1+x)^{1/x}}-1\right)$$ Use the MaLaurin expansion $$(1+x)^{1/x}=e(1-x/2+11x^2/24+...)$$ Then $$\lim_{x \rightarrow 0} \frac{1}{x} [(1-x/2+..)^{-1}-1]= \lim_{x \rightarrow 0} \frac{1}{x}(1+\frac{x}{2}-1]=\frac{1}{2}$$

Answer 4

Let $$a_n=\left(1+\frac{1}{n}\right)^n\implies \log(a_n)=n \log\left(1+\frac{1}{n}\right)$$ Now, using Taylor series $$\log(a_n)=n \left(\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{3 n^3}+O\left(\frac{1}{n^4}\right)\right)=1-\frac{1}{2 n}+\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ Continue with Taylor series $$a_n=e^{\log(a_n)}=e\left(1-\frac{1}{2 n}+\frac{11}{24 n^2}\right)+O\left(\frac{1}{n^3}\right)$$ $$n\left(\frac{e}{\left(1+\frac{1}{n}\right)^n}-1\right)=n\left(\frac{e}{e\left(1-\frac{1}{2 n}+\frac{11}{24 n^2}\right)+O\left(\frac{1}{n^3}\right)}-1\right)$$ Now, long division to get $$n \left(\frac{1}{2 n}-\frac{5}{24 n^2}+O\left(\frac{1}{n^3}\right) \right)=\frac{1}{2 }-\frac{5}{24 n}+O\left(\frac{1}{n^2}\right)$$ which gives the limit and also how it is approached.

Using your pocket calculator for $n=10$, the "exact" value would be $0.4802$ while the truncated expression given above leads to $\frac{23}{48}\approx 0.4792$.