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Picking up from where this question left off...

I have my own take on this proof. I started by showing, by induction, that the sequence is monotonically increasing, and then, also by induction, I showed that the sequence is bounded above by $ x_n<1+\sqrt{a}$. (I can include this if need be) But I want to know if this sequence actually converges to $1+\sqrt{a}$. I know that this sequence is bounded above by this, but I don't know if it is the lease upper bound. Would anyone have any intuition as to prove that, or even where to begin?

Thanks in advance.

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dovedevic
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1 Answers1

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If you know the sequence is bounded and monotonic, then it certainly converges. Now, since you have a relation between $x_{n+1}$ and $x_n$ (that is expressed via a continuous function), you can use a little trick to 'brute force' the value of the limit.

Let $L = \lim_{n\to\infty}x_n$. Then $$L=\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}\sqrt{\alpha +x_n}=\sqrt{\lim_{n\to\infty}(\alpha+x_n)}=\sqrt{\alpha+L}$$

so that $L^2=\alpha+L$, or $L^2-L-\alpha=0$. Then $$L=\frac{1\pm \sqrt{1+4\alpha}}{2}$$

Of course, we know $L>0$, so we pick the positive sign.

Fimpellizzeri
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  • So the brute force part is just reapplying the limit recursively into $x_n$? If so, wow, that makes things very easy! – dovedevic Sep 02 '16 at 00:25
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    Yes! We can do this because the recursion is given by a continuous expression (this justifies the third equality in $L = \dots = \sqrt{\alpha + L}$). Put another way, this kind of recursion makes progressively smaller changes to our $x_n$ iterates (it's converging!), and in the limit it makes no change at all. – Fimpellizzeri Sep 02 '16 at 00:28
  • Fascinating. Thank you! I will have to leave an exercise for myself why that is true. Thank you thank you. – dovedevic Sep 02 '16 at 00:30
  • Looking back at the original question, I see why I got confused with his/her proof, the solution never seemed to explicitly state your step/reasoning above, or why. – dovedevic Sep 02 '16 at 00:33