<s>[INST] 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.
 [/INST] 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:
$$ a_N \leq a_1 \prod_{k\geq 0}\left(1+\frac{1}{4\cdot 2^k+3}\right)=\frac{16}{7}\prod_{k\geq 1}\left(1+\frac{1}{4\cdot 2^k+3}\right)\leq \frac{16}{7}\prod_{k\geq 1}\frac{1+\frac{1}{2^{k+1}}}{1+\frac{1}{2^{k+2}}}=\frac{20}{7}.$$
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