# prove that if $1 \le m \le n$ then $\left(1+\frac1n\right)^m < 1+\frac mn + \left(\frac mn\right)^2$

We know that $\dfrac1n>0$, $\dfrac1n<\dfrac mn<1$, $\left(\dfrac1n+1\right)^m < \left(\dfrac mn+1\right)^m$, so we have to prove that $\left(\dfrac mn+1\right)^m<\dfrac mn+1+\left(\dfrac mn\right)^2$.

I tried using Newton's binomial expression of $\left(\dfrac mn+1\right)^m$ and calculating the difference, but I didn't get there.

• Newton with (1+1/n)^m does not work? – Jean-François Gagnon Sep 9 '17 at 6:25
• the difference did not prove it – Dhia Othmani Sep 9 '17 at 6:28
• My comment does not expand the same binomial expression. – Jean-François Gagnon Sep 9 '17 at 6:29

For integer $m$ we can use induction.
Indeed, for $m=1$ we have $$1+\frac{1}{n}<1+\frac{1}{n}+\frac{1}{n^2}.$$
Let $$\left(1+\frac{1}{n}\right)^m<1+\frac{m}{n}+\frac{m^2}{n^2}.$$ Thus, $$\left(1+\frac{1}{n}\right)^{m+1}<\left(1+\frac{m}{n}+\frac{m^2}{n^2}\right)\left(1+\frac{1}{n}\right).$$ Thus, it remains to prove that $$\left(1+\frac{m}{n}+\frac{m^2}{n^2}\right)\left(1+\frac{1}{n}\right)<1+\frac{m+1}{n}+\frac{(m+1)^2}{n^2}$$ or $$\frac{2m+1}{n^2}>\frac{m^2}{n^3},$$ which is true because $n\geq m$.
Finally, we need to check what happens for $m=n$: $$\left(1+\frac{1}{n}\right)^n<1+1+1$$ or
$$\left(1+\frac{1}{n}\right)^n<3,$$ which is known.