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

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 $\frac{23}{48}\approx 0.4792$.
