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Let $a_n := (1+1/n)^n$ for $n \in \mathbb{N}$

How can one prove that $a_{n+1} \geq a_n$ for all $n \in \mathbb{N}$ with Bernoulli's inequality?

I know that the inequality states that $(1+x)^r \geq 1+rx$ for every integer $r \geq 0$ and every real number $x \geq -2$. And if the exponent $r$ is even, then the inequality is valid for all real numbers x.

So I have to use induction and for $n=1$ we would get $(1+1/1)^1 = (1+1/(1+1))^2$, and that would give $2 < 2,25$. But wouldn't that imply that all numbers $> 1$ would make $a_{n+1} > a_n$? At which case is it equal? Can someone show me where I can find an induction proof for this?

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  • $\begingroup$ There is never equality. Note $a_n \to e$ $\endgroup$ – Henry Nov 14 '19 at 8:42
  • $\begingroup$ @Henry So $a_{n+1} > a_n$ for all $n \in \mathbb{N}$? $\endgroup$ – Ramanujan Taylor Nov 14 '19 at 8:44
  • $\begingroup$ Yes $\,\,\,\,\,\,\,\,$ $\endgroup$ – Henry Nov 14 '19 at 8:45
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    $\begingroup$ In particular there is an answer (among many other answers) that user Bernoulli's inequality $\endgroup$ – Maximilian Janisch Nov 14 '19 at 8:58
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$$ a_{n+1}\ge a_n \quad\Longleftrightarrow\quad \frac{(n+2)^{n+1}}{(n+1)^{n+1}}\ge\frac{(n+1)^n}{n^n} \quad\Longleftrightarrow\quad \frac{n+2}{n+1}\ge\left(\frac{n^2+2n+1}{n^2+2n}\right)^n \\ \quad\Longleftrightarrow\quad \left(1-\frac{1}{n^2+2n+1}\right)^n\ge 1-\frac{1}{n+2} $$ The last inequality holds since $$ \left(1-\frac{1}{n^2+2n+1}\right)^n\ge 1-\frac{n}{n^2+2n+1} $$ and $$ \frac{1}{n+2}\ge\frac{n}{n^2+2n+1} $$

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