Prove that the sequence $\left( 1+\frac{1}{n} \right)^n $ when $n$ $\to$ $\infty$ converges to a number between $2.7$ and $2.8$ using binomial theorem Prove that the sequence $\left( 1+\frac{1}{n} \right)^n $ when $n$ $\to$ $\infty$ converges to a number between $2.7$ and $2.8$ using binomial theorem.
My try
I need help proving that the sequence converges to a number between $2.7$ and  $2.8$ (i know its $e$), but the problem states that i need to use binomial theorem to prove it.
I tried to expand the sequence and i prove that is increasing, and i tried to use the sequence $\left( 1+\frac{1}{n} \right)^{n+1} $ that is decreasing but i can't prove it with this theorem. Any hints?
 A: For the lower bound, just expand the binomial formula till it is greater than $2.7$ because we are summing positive terms, so any partial sum is smaller than the whole.
Unfortunately this is quite tedious since we need to go up to $k=4$ to reach this goal.
$\displaystyle \left(1+\frac 1n\right)^n=\sum_{k=0}^n{n\choose k}n^{-k}\ge \sum_{k=0}^4{n\choose k}n^{-k}=\underbrace{\dfrac{65}{24}}_{\approx \ 2.70833}\underbrace{-\dfrac{5}{4n}+\dfrac{19}{24n^2}-\dfrac{1}{4n^3}}_{\ll 1}>2.7\quad$ for $n$ large enough.
For the upper bound notice that $n!=(n-k)!\times\underbrace{(n-k+1)}_{\le n}\cdots\underbrace{(n-1)}_{\le n}\underbrace{(n)}_{\le n}\le (n-k)!\ n^k$
Thus we get $\quad\displaystyle \sum_{k=0}^n{n\choose k}n^{-k}\le\sum_{k=0}^n \dfrac 1{k!}$
Once again we have to expand the partial sum until we get something less than $2.8$
We will bound the resulting sum using $k!=\underbrace{1.2.3.4}_{>2^4}.\underbrace{5}_{>2}.\underbrace{6}_{>2}\cdots \underbrace{k}_{>2}>2^k$
Applying gives $\quad\displaystyle \sum\limits_{k=0}^{\infty} \dfrac 1{k!}=\dfrac 11+\dfrac 11+\dfrac 12+\dfrac 16+\sum\limits_{k=4}^{\infty} \dfrac 1{k!}\le \dfrac 83+\sum\limits_{k=4}^{\infty} \dfrac 1{2^k}\le \dfrac 83+\dfrac 18\le \underbrace{\dfrac{67}{24}}_{\approx\ 2.7916}< 2.8$
