# Proof $a_{n+1} \geq a_n$ with $a_n := (1+1/n)^n$ for $n \in \mathbb{N}$ [duplicate]

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?

• There is never equality. Note $a_n \to e$ – Henry Nov 14 '19 at 8:42
• @Henry So $a_{n+1} > a_n$ for all $n \in \mathbb{N}$? – Ramanujan Taylor Nov 14 '19 at 8:44
• Yes $\,\,\,\,\,\,\,\,$ – Henry Nov 14 '19 at 8:45
• In particular there is an answer (among many other answers) that user Bernoulli's inequality – Maximilian Janisch Nov 14 '19 at 8:58

## 1 Answer

$$a_{n+1}\ge a_n \quad\Longleftrightarrow\quad \frac{(n+2)^{n+1}}{(n+1)^{n+1}}\ge\frac{(n+1)^n}{n^n} \quad\Longleftrightarrow\quad \frac{n+2}{n+1}\ge\left(\frac{n^2+2n+1}{n^2+2n}\right)^n \\ \quad\Longleftrightarrow\quad \left(1-\frac{1}{n^2+2n+1}\right)^n\ge 1-\frac{1}{n+2}$$ The last inequality holds since $$\left(1-\frac{1}{n^2+2n+1}\right)^n\ge 1-\frac{n}{n^2+2n+1}$$ and $$\frac{1}{n+2}\ge\frac{n}{n^2+2n+1}$$