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.
