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Consider this inequality

$$\left(1-\dfrac{1}{n^2}\right)^n\times\left(1+\dfrac 1n\right)<1$$ which is meant to be valid for any nonzero natural number $n$.

It is asked to prove it by induction. I haven't made any significant progress after many tries.

Any advice is welcome.

Note :

A way I encountered which looks nicer and more straightforward is to notice that $1-\dfrac{1}{n^2}<1-\dfrac{1}{n^2+k}$ for $k\in \{1,\cdots,n\}$

Multiplying side by side those $n$ inequalities, a nice telescoping takes place to give exaclty the sought result.

user21820
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Oussama Sarih
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3 Answers3

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Actually, to prove $$\left(1-\dfrac{1}{n^2}\right)^n\times\left(1+\dfrac 1n\right)=\left(1-\frac{1}{n}\right)^n\left(1+\frac{1}{n}\right)^{1+n}<1$$ It suffices to prove $$\left(1+\frac{1}{n}\right)^{1+n}\le \left(1+\frac{1}{1-n}\right)^n$$ which is obvious since $\left(1+\frac{1}{n}\right)^{1+n} $ is a decreasing sequence.

robjohn
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J.Guo
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Applying Bernoulli's Inequality to the reciprocal, we get $$ \begin{align} \left(1+\frac1{n^2-1}\right)^n\left(1-\frac1{n+1}\right) &\ge\left(1+\frac{n}{n^2-1}\right)\left(1-\frac1{n+1}\right)\\ &=\frac{n^3+n^2-n}{n^3+n^2-n-1} \end{align} $$ Therefore, we get $$ \left(1-\frac1{n^2}\right)^n\left(1+\frac1n\right) \le1-\frac1{n^3+n^2-n} $$

robjohn
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The inequality $\left(1-\dfrac{1}{n^2}\right)^n\cdot\left(1+\dfrac 1n\right) =\left(1-\dfrac{1}{n}\right)^n\left(1+\dfrac{1}{n}\right)^{n+1} =\dfrac{\left(1+\dfrac{1}{n}\right)^{n+1}}{\left(1+\dfrac{1}{n-1}\right)^{n}}<1$ is equivalent to the sequence $\left\{\left(1+\dfrac{1}{n}\right)^{n+1}\right\}$ is strictly decreasing.

So we will show that：sequence $\left\{\left(1+\dfrac{1}{n}\right)^{n+1}\right\}$ is strictly decreasing. By AG-MG inequality $\dfrac{1}{\left(1+\dfrac{1}{n}\right)^{n+1}}=\left(\dfrac{n}{n+1}\right)^{n+1} =1\cdot\dfrac{n}{n+1}\cdot\dfrac{n}{n+1}\cdots\dfrac{n}{n+1} <\left(\dfrac{n+1}{n+2}\right)^{n+2}$ $=\left(\dfrac{n+1}{n+2}\right)^{n+2}=\dfrac{1}{\left(1+\dfrac{1}{n+1}\right)^{n+2}}.$

Hence $$\left(1+\dfrac{1}{n}\right)^{n+1}>\left(1+\dfrac{1}{n+1}\right)^{n+2}.$$

Riemann
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  • (+1) this is a fleshing out of the idea in J.Guo's answer. – robjohn Nov 03 '18 at 14:18
  • @robjohn It is well know that the sequence $\left{\left(1+\dfrac{1}{n}\right)^{n}\right}$ is strictly increasing tending to $e$ and the sequence $ \left{\left(1+\dfrac{1}{n}\right)^{n+1}\right}$ is strictly decreasing tending to $e$ in the standard Calculus textbook. – Riemann Nov 04 '18 at 04:38
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    "well-known" is a relative term. If these are well-known to everyone who reads this answer, then this answer provides nothing more than J.Guo's answer and my upvote was misplaced. However, I think the AM-GM proof here may be new to some readers and, therefore, is worth the upvote. – robjohn Nov 04 '18 at 04:56
  • Thank you for your comment ! – Riemann Nov 04 '18 at 09:33