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Investigating this sequence led me to the following conclusions:

  • if $a_{1}\geq 2$ then $a_{n}\geq 2$ and $a_{n}$ is decreasing, thus it converges and is bounded.
  • if $-1\lt a_{1} \lt 2$ then $a_{n}$ is increasing, thus it converges and is bounded.
  • if $a_{1}=-1$ or $a_{1}=2$ the sequence is constant and is bounded.
  • if $a_{1}\lt -1$ then for some $k,~a_{k}\gt 0$ and then we're in one of the previous options.

But this seems overly complicated and very long to prove.

I thought of trying the following argument: define $f(x)=\frac{4x+2}{x+3}$. We see that $\displaystyle \lim_{x\to \infty}f(x)=4$ and $f(x)$ is continuous in the interval $[0, \infty)$ therefore it's bounded there. Then we show that if $a_{1}<0$ then $\exists N,~\forall n\gt N,~a_{n}>0$ which gives us $a_{n+1}=f(a_{n})$ is bounded.

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What about a_1 = -11/7? There are an infinite values for a_1 for which a_n = -3 ultimately. – Aryabhata Nov 21 '10 at 21:02
to make this complete I guess you'd have to specify a value for $a_{n+1}$ for when $a_n = -3$. – Zarrax Nov 21 '10 at 23:14
@Zari: Yes. I suggest $a_{n+1} = 2$ if $a_{n} = -3$ :-) – Aryabhata Nov 22 '10 at 2:22
A natural move would be to complete $\mathbb R$ by one (!) point at infinity and for $a_n=3$ define $a_{n+1}:=\infty$, $a_{n+2}:=4$. – Christian Blatter Nov 22 '10 at 10:33
up vote 2 down vote accepted

You can formalize what you're doing by showing that $-1 \leq a_{n+1} < 4$ whenever $a_n \geq -1$. So once you get an $n$ such that $a_n \geq -1$ you're done.

If some $a_n < -3$, your recursion gives $a_{n+1} > -1$ and you're in the above situation. The only remaining option is that $-3 < a_n < -1$ for all $n$... but then the sequence is still bounded. So you're done.

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This is what I tried but couldn't show it. For example: if $a_{1}\lt 3$, how do I show that there's some $a_{k}\gt -1$? – daniel.jackson Nov 22 '10 at 20:16
if $a_n < -3$ (I assume that's what you mean), then the condition that $a_{n+1} > -1$ is equivalent to ${4a_n + 2 \over a_n + 3} > -1$, which in turn is equivalent to $4a_n + 2 < -(a_n + 3)$, which in turn is equivalent to $5a_n < -5$, which is true since $a_n < -3$. So if any $a_n < -3$ then $a_{n+1} > -1$. – Zarrax Nov 23 '10 at 0:37

Write $a_n={2+b_n \over 1-b_n}$. Then $b_{n+1}={3\over5} b_n$. This implies $\lim_{n\to\infty} b_n=0$ and $\lim_{n\to\infty} a_n=2$, unless $b_n=1$ for some $n\ge0$. This is the case if $b_0=(5/3)^k$ for some $k\ge0$. In addition there is the fixed point $a_0=a_1=\ldots=-1$ which is not covered by this "Ansatz".

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Wow, your first line is a very clever trick; how did you come up with it? – ShreevatsaR Nov 22 '10 at 12:35
@ShreevatsaR: The Moebius transform $f(x):={4x +2 \over x+3}$ is at stake here. It has the two fixed points 2 and -1. So I introduce the new coordinate variable $t:={x-2\over x+1}$. In terms of $t$ the map $f$ has the form $\tilde f(t):={3\over 5} t$. – Christian Blatter Nov 22 '10 at 13:21

What about $a_1 = -3$? The function don't even exists there.

Anyway, your argument is valid: if $a_n = f(n)$ and $\lim_{x\to \infty} f(x) = L$ then $\lim_{n\to \infty} a_n = L$.

You never care about the first terms of the sequence, because when you've convergence to the infinity, you can just pick a maximum value and set it as limit. For example, if your sequence converges to 4, you can set a bound as $\max (4, a_1, a_2, ..., a_k)$ for any k you like.

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Your second sentence, though correct, is misdirecting. One would be tempted to think $f(x)=\frac{4x+2}{x+3}$, but taking the limit as $x\rightarrow \infty$ gives 4, while the limit of the $a_n$ is 2. – Ross Millikan Nov 21 '10 at 15:31

Your last argument is sound. If you know that the terms are bounded over $[0,\infty)$, (or any interval, for that matter) that iteration doesn't take it out of that interval, and that all starting values eventually enter that interval, the terms are bounded. As AkiRoss says, prepending terms cannot change the bounded nature, it only changes the bound.

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