# Divergence of a recursive sequence

If $(x_n)$ isthe sequence defined by $x_1=\frac{1}{2}$ and $x_{n+1}=\sqrt{x_n^2 +x_n +1}$, show that $\lim x_n = \infty$

Ive tried a couple of things but none of them helped. Ive tried to suppose, by contradiction, that the sequence is bounded, find a lower sequence that goes to infinite and the definition.

• Show by induction that $x_n > \sqrt{n}$ for $n\geq 3$. This is easy as $x_{n+1} = \sqrt{x_n^2+x_n + 1} > \sqrt{n + 1 +\sqrt{n}} > \sqrt{n+1}$. – Winther Aug 11 '14 at 22:00

$$x_{n+1}=\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}>x_n+\frac{1}{2}$$ This implies that $\frac{1}{2}n\leq x_n$ so $x_n$ diverges to $\infty$

• I've tried it, but I don't see hpw this helps. – Giiovanna Aug 11 '14 at 22:02

The sequence is increasing. If it were bounded above, it would have a limit $L$. Then $$L=\lim_{n\to\infty} x_{n+1}=\lim_{n\to\infty} \sqrt{x_n^2+x_n+1}=\sqrt{L^2+L+1}$$ would give $L=\sqrt{L^2+L+1}$, which is impossible, since $L\gt 0$.

• I dont get the equality – Giiovanna Aug 11 '14 at 22:01
• Ok, and the fact that the sequence is increasinb guarantee that it diverges to $\infty$, right? – Giiovanna Aug 11 '14 at 22:05
• If the limit is $L$, then $\lim_{b\to\infty} x_{n+1}=L$. But $x_{n+1}=\sqrt{x_n^2+x_n+1}$, and because $\sqrt{t^2+t+1}$ is a continuous function, $\lim_{n\to\infty}\sqrt{x_n^2+x_n+1}=L^2+L+1$. – André Nicolas Aug 11 '14 at 22:06
• I got the part of why it cannot converge to a real number, but I also have to prove that it diverges to infinite. But I think that it is guaranted by the fact that the sequence is increasing. – Giiovanna Aug 11 '14 at 22:07
• The fact that the sequence is increasing guarantees that if it is bounded, it has a (finite) limit. But we have shown that any limit has to satisfy $L=\sqrt{L^2+L+1}$, i.e. $L^2=L^2+L+1$. This is impossible, so the sequence cannot be bounded. And yes, you are right, increasing and not bounded above implies divergence to infinity. – André Nicolas Aug 11 '14 at 22:08