proof 2I'm a beginner in demonstrations and I'm trying to prove that $$2< \lim_{n\to\infty} \left(1+\frac1n\right)^n< 3.$$
I proved $$2< \lim_{n\to\infty} \left(1+\frac1n\right)^n,$$ by Newton's binomial, but can't prove that $$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n< 3.$$
Can you give me a light?
(I know this converges to "$e$", but I want to prove it without this.)
 A: HINT
Use that, by Binomial theorem
$$\left(1+\frac1n\right)^n\le\sum_{k=0}^n \frac1{k!}=1+\sum_{k=1}^n \frac1{k!}\le 1+\sum_{k=1}^n \frac1{2^{k-1}}=1+\sum_{k=0}^{n-1} \frac1{2^k}$$
and note also that by geometric series
$$\sum_{k=0}^{n-1} \frac1{2^k}=\frac{1-\frac1{2^n}}{1-\frac12}=2\left(1-\frac1{2^n}\right)<2$$
A: Use $AM-GM$ first, to show that $(1 + \frac 1n)^n \leq (1 + \frac 1{n+1})^{n+1}$ for all $n \geq 1$. To do this, perform $AM-GM$ on $n$ copies of $(1 + \frac 1n)$ and the number $1$. The arithmetic mean is $\frac{1 + n(1 + \frac 1n)}{n+1} = \frac{n+2}{n+1} = 1 + \frac 1{n+1}$ and the geometric mean is $\sqrt[n+1] {(1 + \frac 1n)^n}$, so $AM \geq GM$ gives the desired result.
Since $n=1$ gives $(1 + \frac 1n)^n = 2$, we get that every term of the sequence is greater than or equal to $2$, so the limit is greater than or equal to $2$.
For the other bound, I  want you to complete the steps I give you, because the proof is elementary, but it would not be good for you if I just do it and give it to you.We will prove that $(1 + \frac 1n)^n < 3$ for all $n$. This will show that the limit must also be less than (or equal to) $3$.


*

*First, binomially expand $(1 + \frac 1n)^n$. You will get a sum, of some powers of $n$ times some binomial coefficients which depend on $n$.

*Take all the negative powers of $n$ to the numerator, so you have some expression looking like $1 + \frac {c_1}{1!} + \frac{c_2}{2!} + ... + \frac{c_n}{n!}$,where $c_i$ are expressions which will depend on $n$.

*See that when you do this, each $c_i \leq 1$. So you get an inequality, that $(1 + \frac 1n)^{n} \leq 1 + \frac 1{1!} + \frac 1{2!} + ... + \frac 1{n!}$.

*Use the fact that $n! > 2^{n-1}$ to bound this sum above by a geometric series. 

*Sum the geometric series to get the result.
