Computing $ \lim _{n \rightarrow \infty}\left[n-\frac{n}{e}\left(1+\frac{1}{n}\right)^{n}\right]$

Question

$$ \lim _{n \rightarrow \infty}\left[n-\frac{n}{e}\left(1+\frac{1}{n}\right)^{n}\right] \text { equals }\_\_\_\_ $$

I tried to expand the term in power using binomial theorem but still could not obtain the limit.

Answer 1

One has $$n - \frac{n}{e} \left( 1 + \frac{1}{n}\right)^n = n - \frac{n}{e} \exp \left( n \ln \left( 1 + \frac{1}{n}\right)\right) =n - \frac{n}{e} \exp \left( n \left(\frac{1}{n}- \frac{1}{2n^2} + o \left( \frac{1}{n^2}\right)\right)\right) $$

so $$n - \frac{n}{e} \left( 1 + \frac{1}{n}\right)^n = n - \frac{n}{e} \exp \left( 1 -\frac{1}{2n} + o \left( \frac{1}{n}\right)\right) = n - n \left( 1 - \frac{1}{2n} + o \left( \frac{1}{n}\right)\right) $$

so $$n - \frac{n}{e} \left( 1 + \frac{1}{n}\right)^n = \frac{1}{2} + o(1).$$

Therefore the limit is $$\frac{1}{2}$$

Answer 2

The problem is that you cannot use Binomial Theorem because you cannot control the behaviour of $\binom{n}{k}\frac{1}{n^k}$ when $n$ and $k$ increase towards infinity. One must expand $(1+\frac{1}{n})^n$ using Taylor Series, it's more straight to the point.