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Let (*) be $ \left(1+\frac{1}{n}\right)\left(1-\frac{1}{n^2}\right)^n < 1, ∀n∈ℕ$

I have tried many ways to get to (*)

We have $\left(1-\frac{1}{n^2}\right)^n <1$ and $-\left(1+\frac{1}{n}\right) < -1 <1$

How can you infer (*) from these two last inequalities?

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Proof: We know that $a_n=(1+\frac{x}{n})^n$ is a strictly increasing sequence for $x\geq -1$ and it is bounded above to $e^x$.Letting $x=\frac{-1}{n}$, we get: $$(1-\frac{1}{n^2})^n < e^{-1/n}$$

Therefore, $$(1+\frac{1}{n})(1-\frac{1}{n^2})^n < (1+\frac{1}{n})\times e^{-1/n}$$

But, it is a well-known fact that for all real values $x \in \mathbb{R}$, we have $e^x \geq 1+x$ (a proof is available here). Hence,

$$e^{1/n}\geq 1+\frac{1}{n} \implies (1+\frac{1}{n})\times e^{-1/n} \leq 1$$

And we obtain: $$(1+\frac{1}{n})(1-\frac{1}{n^2})^n < 1$$ Q.E.D.

Addendum: To know why $a_n=(1+\frac{x}{n})^n$ is bounded above to $e^x$, note that $a_n$ is strictly increasing and $\lim_{n\to \infty}a_n=e^x$. So, it has to be bounded from the above by $e^x$. To know why $a_n$ is strictly increasing, from the AM-GM inequality we have:

$$\sqrt[n+1]{x_{1}x_{2}\cdots x_{n+1}}<\frac{x_{1}+x_{2}+\ldots+x_{n+1}}{n+1}%$$ Let $x_1=1$ and $x_2=\cdots=x_{n+1}=1+\frac{x}{n}$ $$ \left( 1+\frac{x}{n}\right) ^{\frac{n}{n+1}}<\frac{1+n\left( 1+\frac{x}{n}\right) }{n+1}=1+\frac{x}{n+1}% $$

$$a_n=\left( 1+\frac{x}{n}\right)^{n}<\left(1+\frac{x}{n+1}\right)^{n+1}=a_{n+1} $$

Remark: It's irrelevant but interesting to see that using the Bernoulli's inequality we get a lower bound for the expression as follows:

$$(1-\frac{1}{n^2})^n \geq 1-\frac{1}{n}$$

$$(1+1/n)(1-\frac{1}{n^2})^n \geq \frac{n+1}{n}\frac{n-1}{n}=\frac{n^2-1}{n^2}$$

Therefore,

$$\frac{n^2-1}{n^2}<(1+\frac{1}{n})(1-\frac{1}{n^2})^n$$

Therefore, as you could see, proving that it's smaller than $1$ is actually harder than I initially thought because the expression quickly gets very close to $1$. ;-)

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  • $\begingroup$ can you elaborate more on how the last one is obtained? $\endgroup$ – user492220 Nov 14 '17 at 2:08
  • $\begingroup$ I can't see it really leads to obtain that inequality. Can you do it manually? $\endgroup$ – user492220 Nov 14 '17 at 3:00
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    $\begingroup$ Macavity, Yes, that cannot be done, because we have 1+(1/n) > 1 not < 1 $\endgroup$ – user492220 Nov 14 '17 at 3:36
  • $\begingroup$ how would you prove that statement? $\endgroup$ – user492220 Nov 14 '17 at 3:40
  • $\begingroup$ I have already settled it down with stating these inequalities, my main goal is how you get the wanted inequality? $\endgroup$ – user492220 Nov 14 '17 at 3:42
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While you already have an answer, here is another way: $n=1$ is trivial and for $n> 1$, by AM-GM, $$\sqrt[n+1]{\left(1+\frac1n \right) \left(1 - \frac1{n^2} \right)^n} < \frac1{n+1}\left[\left(1+\frac1n\right) + n\left(1 - \frac1{n^2} \right) \right] = 1$$ Note equality is never possible as $1+\frac1n \neq 1-\frac1{n^2}$, so the inequality is strict.

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