Show that $2 < (1+\frac{1}{n})^{n}< 3$ without using log or binommial coefficient $2 <(1+\frac{1}{n})^{n}< 3$
Is it possible to show the inequality without using binommial coefficients thus only by induction? The leftinequality can be shown using bernoulli inequality.
 A: Let $a_n=\left(1+\frac{1}{n}\right)^n$. For any $n>1$ the inequality $a_n>2$ is trivial. We have $a_{n+1}>a_n$ by AM-GM:
$$\sqrt[n+1]{1\cdot a_n} = \text{GM}\left(1,1+\tfrac{1}{n},\ldots,1+\tfrac{1}{n}\right) \stackrel{\text{AM-GM}}{<} \tfrac{1}{n+1}\left[1+n\cdot\left(1+\tfrac{1}{n}\right)\right] = 1+\tfrac{1}{n+1}.$$
Additionally:
$$ \frac{a_{2n}}{a_n} = \left(1+\frac{1}{4n(n+1)}\right)^n \leq \frac{1}{1-\frac{1}{4(n+1)}} = 1+\frac{1}{4n+3}$$
hence for any $N\geq 1$ we have:
$$ a_N \leq a_1 \prod_{k\geq 0}\left(1+\frac{1}{4\cdot 2^k+3}\right)=\frac{16}{7}\prod_{k\geq 1}\left(1+\frac{1}{4\cdot 2^k+3}\right)\leq \frac{16}{7}\prod_{k\geq 1}\frac{1+\frac{1}{2^{k+1}}}{1+\frac{1}{2^{k+2}}}=\frac{20}{7}.$$
A: You have the left inequality already (Bernoulli).
The right inequality can be expressed as 
$$
{n} \log (1+\frac{1}{n}) < \log 3
$$
Now $\log(1+x) < x$  by concavity, which gives $1 < \log 3$ which is true.
A: I have found the answer to my problem in a different forum:
We show that $b_n > b_{n+1}$ with $b_n:= (1+\frac{1}{n})^{n+1}$
Proof:
To show: $\frac{b_n}{b_{n+1}}\geq 1$
$\frac{b_n}{b_{n+1}} = \frac{\left(1+\frac{1}{n}\right)^{n+1}}{\left(1+\frac{1}{n+1}\right)^{n+2}} = \frac{1}{1+\frac{1}{n}} \cdot \left(\frac{1+\frac{1}{n}}{1+\frac{1}{n+1}}\right)^{n+2} = \frac{1}{1+\frac{1}{n}} \cdot \left(1+\frac{1}{n(n+2)}\right)^{n+2}$
Exploiting the bernoulli inequality
$\frac{1}{1+\frac{1}{n}} \cdot \left(1+\frac{1}{n(n+2)}\right)^{n+2} \geq 1$
$b_5 < 3$ and $(1+\frac{1}{n})^n < b_n $
Thus $\forall n\geq 5: (1+\frac{1}{n})^n < 3$ 
