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Suppose $a_0 = \sqrt2$, $a_1 = 3$ and $a_{n+1} = 2a_n^2 −1$ for all natural numbers $n ≥ 1$. Define $b_n = 2^na_0a_1 ···a_{n−1}$ for all $n ≥ 1$. Show that $\lim_{n\to\infty} \frac{a_n}{b_n} = 1.$

My Proof: $$a_n = 2a_{n-1}^2-1 = 2a_{n-1}^2 -2 +1 = 2(a_{n-1}^2 -1)+1 = 2(a_{n-1}-1)(a_{n-1}+1)+1$$ $$ = 2(2a_{n-2}^2 -1-1)(a_{n-1} +1)+1 = 2^2(a_{n-2}-1)(a_{n-2}+1)(a_{n-1}+1)+1$$ $$= ...= 2^{n-1}(a_1 -1)(a_1+1)(a_2+1)....(a_{n-1}+1)+1$$ $$=2^n(a_1+1)(a_2+1)....(a_{n-1}+1)+1\\$$

Since $a_n$ > 1, it is obvious that $a_0a_1 ···a_{n−1}$ diverges to $\infty$.

Using $a_{n+1} + 1 = 2a_n^2$,

$$\lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty}\frac{2^n(a_1+1)(a_2+1)....(a_{n-1}+1)+1}{2^na_0a_1 ···a_{n−1}}$$ $$=\lim_{n\to\infty}\frac{(a_1+1)(a_2+1)....(a_{n-1}+1)+1}{a_0a_1 ···a_{n−1}}$$ $$=\lim_{n\to\infty}\frac{(2a_0^2)(2a_1^2)....(2a_{n-2}^2)}{a_0a_1 ···a_{n−1}} + \frac{1}{a_0a_1 ···a_{n−1}}$$ $$=\lim_{n\to\infty}\frac{2^{n-1}(a_0^2a_1^2....a_{n-2}^2)}{a_0a_1 ···a_{n−1}}$$ $$=\lim_{n\to\infty}\frac{2^{n-1}(a_0a_1....a_{n-2})}{a_{n−1}}$$ $$=\lim_{n\to\infty} \frac{b_{n-1}}{a_{n-1}}$$ $$=\lim_{n\to\infty} \frac{1}{\frac{a_{n-1}}{b_{n-1}}}$$ $$=\lim_{n\to\infty} \frac{1}{\frac{b_{n-2}}{a_{n-2}}}$$ $$=\lim_{n\to\infty} \frac{a_{n-2}}{b_{n-2}}$$ $$= ... =\lim_{n\to\infty} \frac{a_0}{b_0} = \frac{\sqrt2}{\sqrt2} = 1$$

Is there any simpler way of doing this? Is my method even correct? Help appreciated thanks.

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  • $\begingroup$ The last line is not really rigorous. It would be better to say that $\lim \frac{b_{n-1}}{a_{n-1}} = \lim \frac{b_n}{a_n}$, and that $\lim \frac{a_n}{b_n} = \lim \frac {b_n}{a_n}$ implies $ \lim \frac{a_n^2}{b_n^2} = 1$, but this is not $-1$, because we have a positive sequence. Also you must guarantee that the limit exists before calculating it. $\endgroup$ – user135508 Apr 29 '14 at 13:32
  • $\begingroup$ I don't really get what u mean. Can u explain more please thanks $\endgroup$ – user10024395 Apr 29 '14 at 13:36
  • $\begingroup$ Consider the following limit evaluation: $\lim 1 = \lim \frac{1}{n} + \cdots \frac{1}{n} (n times) = \lim \frac{1}{n} + \cdots + \lim \frac{1}{n} (n times) = 0 + \cdots + 0 = 0$. So it is illegal to do something "n times" while n is approaching infinity. About the existence, I could provide the similar paradox: $\lim 2^n = 2(\lim 2^{n-1}) = 2(\lim 2^n)$. $x=2x$ implies $x=0$ ! By the way, do you understand why $\lim a_n = \lim a_{n-1}$? $\endgroup$ – user135508 Apr 29 '14 at 13:45
  • $\begingroup$ I think so. Because the shift in index doesn't matter? Btw how can I know when I can spli the limit computation into 2 parts cos for the first step I have already splitter it once. Is that illegal also? I know the example shows that it is illegal but how to know when the splitting is ok when it is not? Thanks $\endgroup$ – user10024395 Apr 29 '14 at 15:11
  • $\begingroup$ @user135508 When I think more about it I think your example might be different because the n in your example is not actually infinity, i mean u can't have infinity multiplied by 0 = 0 can you? it just doesn't feel right. Does it? $\endgroup$ – user10024395 Apr 29 '14 at 15:28
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Note that $$ (2a_{n-1})^2=2(2a_{n-1}^2)=2(a_n+1)=2\frac{a_n^2-1}{a_n-1}= \frac{a_n^2-1}{ a_{n-1}^2-1} $$ Thus $$ b_n=\prod_{k=0}^{n-1}(2a_k)= \sqrt{\prod_{k=0}^{n-1}\frac{a_{k+1}^2-1}{ a_{k}^2-1}}= \sqrt{\frac{a_{n}^2-1}{ a_{0}^2-1}}= \sqrt{a_{n}^2-1}. $$ So $$\dfrac{b_n}{a_n}=\sqrt{1-\dfrac{1}{a_n^2}}$$ and the conclusion follows since $\lim\limits_{n\to\infty}a_n=+\infty$.$\qquad\square$

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  • $\begingroup$ Thanks. Btw is there anything wrong with my method although it is very messy and complicated. $\endgroup$ – user10024395 Apr 29 '14 at 13:34

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