> 
> Let, $a\_1=2$ and for $n\ge 2$, $a\_{n+1}=a\_n^2-a\_n+1$. Prove that the sequence $\{a\_n\}$ is divergent.
> 
> 
> 

It is clear that , $a\_{n+1}-a\_n=(a\_n-1)^2\ge 0$ which implies $a\_{n+1}\ge a\_n.$ So the sequence $\{a\_n\}$ is monotonic increasing. So it is sufficient to show that $\{a\_n\}$ is unbounded above !

If $\{a\_n\}$ is bounded then then $\{a\_n\}$ is convergent, say converges to $l$. i.e., $a\_n \to l$ as $n\to \infty$. Then, from the given relation we have, $l=l^2-l+1$ , which implies $l=1$, contradicts that $a\_1=2$. So we are done !

Any other way to show that the given sequence is unbounded above ?