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Question 8.12 of Apostol's Analysis asks us to prove that $\{a_n\}$, with $a_1 = -3/2$ and $3a_{n+1} = 2 + a_n^3$, converges to $1$. My approach consists on proving that it's increasing and bounded above. I've succesfully proven that it is monotone by induction, but I can't seem to find an upper bound. How does one find an upper bound for such a sequence?, any help?

In this question, the OP proposes a way of proving that $\{a_n\}$ is bounded above by considering the function $f(x) = \frac{1}{3}(2+x^3)$ is increasing in $(-2,1)$ and then evaluating at $1=f(1)$, but I still don't get the point: why is that a valid argument?

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    $\begingroup$ Since the limit is supposed to be $1$, and it is increasing, probably need to show that $1$ is an upper bound. Pretty easy to prove that by induction, too. $\endgroup$ – Thomas Andrews Nov 18 '15 at 23:18
  • $\begingroup$ @ThomasAndrews Should I go at it like trying to prove that for all $n$, $a_n\leq 1$?, should I use induction? $\endgroup$ – Miguelgondu Nov 18 '15 at 23:19
  • $\begingroup$ @hamidkamali at first, I deleted it when I found said question. Then, as I read it, I found out that I didn't understand it correctly. That's precisely why I linked it. $\endgroup$ – Miguelgondu Nov 18 '15 at 23:32
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If $a_n<1$ then also $a_n^3<1$ and thus $$ a_{n+1}<\frac13(2+1)=1. $$

One can even say more in that $$ 1-a_{n+1}=\frac13(1-a_n^3)=\frac13(1-a_n)(a_n^2+a_n+1)=\frac13(1-a_n)((a_n+\tfrac12)^2+\tfrac34) $$ which is contractive as long as $|a_n+\tfrac12|<\sqrt{3-\frac34}=\frac32$.

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  • $\begingroup$ Great answer, thanks! $\endgroup$ – Miguelgondu Nov 18 '15 at 23:39

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