Find $\displaystyle \lim_{n \to \infty} \left[n-\frac{n}{e}\left(1-\frac{1}{n}\right)^n\right]. $
My Approach:
$$ \lim_{n \to \infty} \left[n-\frac{n}{e}\left(1-\frac{1}{n}\right)^n\right] = \lim_{n \to \infty} n \left[1-\frac{1}{e}\left(1-\frac{1}{n}\right)^n\right] = \infty \times 0 = 0$$
I got $\infty$ because $n - \frac{n}{e}\left(1-\frac{1}{n}\right)^n= n\left(1-\frac{\left(1-\frac{1}{n}\right)^n}{e}\right)\to \infty$ because $\left(1-\frac{1}{n}\right)^n \to \frac{1}{e}$
Considering $$a_n=n-\frac{n}{e}\left(1-\frac{1}{n}\right)^n$$ you can do a nice work composing Taylor series. Here are the steps $$A=\left(1-\frac{1}{n}\right)^n\implies \log(A)=n \log\left(1-\frac{1}{n}\right)$$ $$\log(A)=n\left(-\frac{1}{n}-\frac{1}{2 n^2}-\frac{1}{3 n^3}+O\left(\frac{1}{n^4}\right)\right)=-1-\frac{1}{2 n}-\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ $$A=e^{\log(A)}=\frac{1}{e}-\frac{1}{2 e n}-\frac{5}{24 e n^2}+O\left(\frac{1}{n^3}\right)$$ $$a_n=n-\frac n e A=\left(1-\frac{1}{e^2}\right) n+\frac{1}{2 e^2}+\frac{5}{24 e^2 n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit.
But it also gives a shortcut approximation of the value of $a_n$. Take $n=10$; the exact result is $$a_{10}=10-\frac{3486784401}{1000000000 e}\approx 8.71728$$ while the above truncated expression gives $$a_{10}\sim 10-\frac{455}{48 e^2}\approx 8.71713$$
