1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ By induction you can show that $x_{n+1} \leq 2a$ if $a \leq 1/4$ $\endgroup$ – echzhen Nov 21 '16 at 14:42
1
$\begingroup$

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 $(x_n)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.