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The question is:

$0\leq\alpha\leq \frac{1}{4}$, We'll define $a_1=\alpha$, $a_{n+1}=\alpha+a_n^2$. Prove $(a_n)_{n=1}^\infty$ is converging and find it's limit.

I'm really confused about it, so I tried taking the private case of $\alpha=\frac{1}{4}$ but I still don't really understand it.
And I'm not really sure where this is going to or what I'm supposed to do with it.

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up vote 2 down vote accepted

Well, there is a way that often works on problems like this. First, suppose that the sequence does converge and find its limit. Hint: if $(a_n) \to x$, then $x = \frac{1}{4} + x^2$.

Next, when you've found the supposed limit value $x$, you'll just need to prove that $a_n - x \to 0$, and the problem will be solved. In the actual solution you can even omit the part where you determine the value of $x$. You can just say out of the blue something like "Let's show that $(a_n)$ converges to <here you name the limit>", and then go on proving that $a_n - x \to 0$. This looks cool and makes people think that you've come up with the limit value using your incredible intuition )).

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Then the crucial ingredient of the proof is indeed to show that $a_n-x\to0$. Your post should include a way to prove that... – Did Nov 28 '12 at 18:28
@did I mean this as a hint. When $x$ is known, proving that $a_n - x$ converges to $0$ becomes way easier than the original problem was. I believe the OP can do it himself. – Dan Shved Nov 28 '12 at 18:29
Sorry but I do not share your optimism. – Did Nov 28 '12 at 18:33
@did Well, if the OP wants it, he can ask me for the rest of the proof and I'll provide it. This isn't optimism. Rather, I think that this is how homework questions should be answered. I must have read something about it in the FAQs, but I cannot find it... – Dan Shved Nov 28 '12 at 18:40
Thank you! That's a nice trick i didn't know... Too bad classes here doesn't bother to teach you how to solve the exercises you get... – Nescio Nov 28 '12 at 19:24

Prove that the sequence is increasing and bounded. Both using induction.
Then find the limit as Dan Shved proposed.

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