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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.

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

Done!

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