Let $a>0.$ Define the sequence $(x_n)_{n\ge 0}$ by $$x_{n+1}=a+x_n^2,\,x_0=0.$$ Find a necessary and sufficient condition such that the sequence is convergent.

If $x_n\to l,$ then by passing the limit and solving the quadratic equation, we get $l=\frac{1 \pm \sqrt{1-4a}}{2}.$ So, necessarily $a\le 1/4.$
I am not able to obtain the sufficient condition. I know that the sequence is increasing which I proved by induction$(x_{n+1}-x_n=x_n^2-x_{n-1}^2).$ By Monotone convergence theorem if I am able to show that the sequence is bounded, I'm done.

  • $\begingroup$ If any $x_n$ exceeds 1, the sequence diverges. Does that help? $\endgroup$ – Matthew Leingang Nov 21 '16 at 14:07
  • $\begingroup$ @MatthewLeingang it seems the op already knows $a\le 1/4$, in particular $a <1$. $\endgroup$ – Adam Hughes Nov 21 '16 at 14:10
  • $\begingroup$ @MatthewLeingang This came to my thoughts, but i realised that still this is only a sufficient condition for divergence of the sequence $\endgroup$ – Bijesh K.S Nov 21 '16 at 14:10
  • $\begingroup$ Ah, I was still thinking about sufficiency. Sorry. Must apply more coffee. $\endgroup$ – Matthew Leingang Nov 21 '16 at 14:13
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    $\begingroup$ By induction you can show that $x_{n+1} \leq 2a$ if $a \leq 1/4$ $\endgroup$ – echzhen Nov 21 '16 at 14:42

This question contains its own answer. $a\leq 1/4$ is the nessessary and sufficient condition.

Indeed, if $a\leq 1/4$, prove by induction that $x_n<1/2$. The base of the induction is trivial. Step of the induction: if $x_n<1/2$, then $x_{n+1}<1/4+(1/2)^2=1/2$.

This proves the boundedness, hence convergence of the sequence $(x_n)$.


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