Prove that $2 < f(n) < 3$ if $n$ is a natural number and $n \ge 2$ and $f(n)=\left (1+\frac{1}{n}\right)^n$.
I've succeded in proving that $f(n)>2$ :
$$ f(n)=\left (1+\frac{1}{n}\right)^n \\ f(n)=\binom{n}{0}+\binom{n}{1}\left(\frac{1}{n}\right)+\binom{n}{2}\left(\frac{1}{n^2}\right)+\binom{n}{3}\left(\frac{1}{n^3}\right)+...+\binom{n}{n}\left(\frac{1}{n^n}\right) $$ Since the first two terms in this series are $1$, $f(n) > 2$. Can anybody give me a few hints on proving the upper bound ? Edit : I already know of the way to do this using limits and am looking for a way to do this using the Binomial Theorem.
By the AM-GM inequality, the sequence $\{a_n\}_{n\geq 1}$ given by $a_n=\left(1+\frac{1}{n}\right)^n$ is increasing:
$$ a_n = 1\cdot\left(1+\frac{1}{n}\right)^n \leq \left(\frac{1+n\cdot\left(1+\frac{1}{n}\right)}{n+1}\right)^{n+1} = a_{n+1} $$ but such sequence is also bounded above: $$ a_n = 2+\sum_{k=2}^{n}\binom{n}{k}\frac{1}{n^k} =2+\sum_{k=2}^{n}\frac{1}{k!}\prod_{j=0}^{k-1}\frac{n-j}{n}\leq 2+\sum_{k\geq 2}\frac{1}{k!}$$ and it is enough to prove that $\sum_{k\geq 2}\frac{1}{k!}<1$, that simply follows from: $$ \sum_{k \geq 2}\frac{1}{k!}<\sum_{k\geq 2}\frac{1}{k(k-1)}=\sum_{k\geq 2}\left(\frac{1}{k-1}-\frac{1}{k}\right)=1.$$