I can not to prove the following inequality $$ \forall \, n\in \mathbb N^* : \quad \left(1- \frac{1}{n^2} \right)^n \left( 1+ \frac{1}{n} \right) < 1$$ I tried to use the recurrence, but I'm stuck. Someone can help me!!!

Remarks: I need proof, to the level of a high school student.

Thanks in advance


1 Answer 1


Check that the inequality holds for $n=1$. For $n\geq 2$ Bernoulli's inequality implies that $$\left(1-\frac{1}{n^2}\right)^{-(n+1)}=\left(1+\frac{1}{n^2-1}\right)^{n+1}>1+\frac{1}{n-1}$$ and so $$\left(1-\frac{1}{n^2}\right)^{n+1} < 1-\frac{1}{n}.$$ Divide both sides by $1-\frac{1}{n}$.

  • $\begingroup$ @ WimC, thanks for your answer, but I need proof, to the level of a high school student $\endgroup$
    – Z. Alfata
    Nov 10, 2015 at 21:56
  • $\begingroup$ Then take the binomial theorem, for $x\ge 0$ you have $(1+x)^n\ge 1+nx$. More was not used. $\endgroup$ Nov 10, 2015 at 23:22
  • $\begingroup$ This is a really nice proof! I've used Bernoulli's Inequality several times in answers I've posted on MSE. It is such an underrated inequality! Well done! +1 $\endgroup$
    – Mark Viola
    Nov 11, 2015 at 0:21

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