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

Question

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?

Answer 1

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