I am given a sequence defined by $$ x_1=2 \ \text{and} \ x_{n+1}=x_n + \dfrac{1}{x_n} \ \text{for} \ n=1,2,\dots $$ I want to show that it is strictly increasing and that $x_n \to +\infty$.
Attempt at a solution:
That it is strictly increasing, I sense I need to use induction. Namely, I want to show that for all $n$, $x_n <x_{n+1}$. This statement is true for $n=1$ and I assume it is true for $n=k$ and I need to show that this implies it is true for $k+1$. Having hit a brickwall with this attempt I need some guidance.
As far as $x_n\to +\infty$ is concerned,I need to show that it is not bounded. For if it is increasing and not bounded this would imply it diverges to infinity. Therefore I have to show that for all $M>0$ there exists a $\tilde{n}=\tilde{n}(M)$ such that $x_{\tilde{n}}>M$.
Is my understanding correct? Any hints on the proof would be welcome.
$x_{n+1}=x_n + \dfrac{1}{x_n}$ implies $x^2_{n+1}=x^2_n +2+ \dfrac{1}{x^2_n}>x^2_n +2$, so we can easily prove $x^2_n\ge2n+2$ by induction: it is true for $n=1$, because $x^2_1=4=2\cdot1+2$. If we know that it's true for $n$, i.e. $x^2_n\ge2n+2$, then $x^2_{n+1}>x^2_n+2\ge2n+2+2=2(n+1)+2$ (induction step).
This means $x_n\ge\sqrt{2n+2}$, since $x_n>0$ (by induction, too). The inequality is strict, except for $n=1$, naturally.