# Divergence of $u_{n+1}=1+\frac{n}{u_n}$

Let $u_n$ be defined by $u_0=1$ and $u_{n+1}=1+\frac{n}{u_n}$.

It can be shown easily that if it has a limit, then it must be $+\infty$.

Does $u_n$ diverge to $+\infty$ ?

What I have tried :

Let $x=u_n$

$U_{n+1}-u_n=1+\frac{n}x-x=f(x)$

$f'(x)=\frac{-n^2}{x^2}-1$

$f$ is sctrictly decreasing, $f\left(\dfrac{-1+\sqrt{1+4n}}2\right)=0$ which is its only $0$

But that diverges towards $\infty$ so I haven't been able to deduce anything from it.

The first terms of the sequence are $u_0 = 1, u_1 = 1, u_2=2, u_3=2, \dots$

I prove by induction that $$\forall n \geq 2,\qquad \ \sqrt{n} < u_n < \sqrt{n}+1.$$

From this follows that $\lim_n u_n = + \infty$.

For $n=2$ we are ok because $\sqrt{2} <2 < \sqrt{2} +1$.

For the inductive step we have

$$u_{n+1} = 1+ \frac{n}{u_n} < 1+ \frac{n}{\sqrt{n}} < 1+ \sqrt{n+1}$$ and $$u_{n+1} = 1+ \frac{n}{u_n} > 1+ \frac{n}{1+ \sqrt{n}} > \sqrt{n+1}$$

Where the last inequality holds because $\forall x >0$ $$1+ \frac{x}{1+ \sqrt{x}} - \sqrt{x+1} >0$$ holds.