<s>[INST] $2 < (1 + \frac{1}{n})^n < 3$ for all $n > 2$ Show by induction that $2 < (1 + \frac{1}{n})^n < 3$ for all natural n > 2.


*

*Induction base. For $n = 3$ the inequality is obviously  true.

*Assume that for $n = k$ inequality is true, than i can prove $(1 + \frac{1}{n+1})^{n +1} < 3$.
$$\frac{(n + 1)^{n}}{n^{n}} < 3 \\
1 < \frac{3n^{n}}{(n+1)^{n}} \\
\frac{(n+2)^{n+1}}{(n+1)^{n+1}} < \frac{3 n^{n} (n + 2)^{n+1} }{(n+1)^{2n +1}}, but \\
\frac{3 n^{n} (n + 2)^{n+1} }{(n+1)^{2n +1}} < 3, then \\
\frac{(n+2)^{n+1}}{(n+1)^{n+1}} < 3
$$
How can i prove, that $(1 + \frac{1}{n+1})^{n +1} > 2$
 [/INST] I'd like to point out an error in the proof for $\left(1+\frac1n\right)^n < 3$.
The step that says $\frac{3 n^{n} (n + 2)^{n+1} }{(n+1)^{2n +1}} < 3$ is wrong (I plugged $n=3$ into the windows calculator and got a term slightly above 3). That term is equal to
$$\frac{3(n^2 + 2n)^n(n+2)}{(n^2+2n+1)^n(n+1)} = 3\frac{(n^2 + 2n)^n}{(n^2+2n+1)^n}\frac{n+2}{n+1}$$
so the product of $3$, a term sligthly below $1$ and a term slightly above $1$. I suspect it will be $>3$ for all cases, but that might not be easy to proof.
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