Let $a_{n}$ be a sequence such that $(a_{n})^{2}=ca_{n-1}$ where ($c>0,a_{1}>0$).Prove that the sequence converges to $c$. My first problem was to find some terms of the sequence to verify that point and show that converges and converges to that point using some convergence criterion. Could someone help me through this problem?
Let $a_{n}$ be a sequence such that $(a_{n})^{2}=ca_{n-1}$ where ($c>0,a_{1}>0$).Prove that $a_n$ converges to $c$.
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1Although my mother language is Spanish (too?) I guess it is necessary to post in English here. I "transledited". – 2012-04-26
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0@Peter: This was also discussed at meta: [What is the site etiquette about (i) asking and (ii) answering questions in a language other than English?](http://meta.math.stackexchange.com/questions/1617/what-is-the-site-etiquette-about-i-asking-and-ii-answering-questions-in-a-la) – 2012-04-26
5 Answers
Let $b_n = \ln(a_n)$. Then your relation is $b_{n+1} = {b_n + \ln(c) \over 2}$. So $b_{n+1} - \ln(c) = {1 \over 2} (b_n - \ln(c))$. From this it is pretty immediate that $b_n $ converges to $\ln(c)$. And then $a_n = e^{b_n}$ converges to $c$ by continuity of $e^x$.
Note that calculating some values gives:
$$\Large \eqalign{ & {a_1} = a \cr & {a_2} = \sqrt c \sqrt a \cr & {a_3} = \sqrt c \root 4 \of c \root 4 \of a \cr & {a_4} = \sqrt c \root 4 \of c \root 8 \of c \root 8 \of a \cr} $$
In general you can prove that
$$\Large {a_n} = {c^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots+ \frac{1}{{{2^{n - 1}}}}}}{a^{\frac{1}{{{2^{n-1}}}}}}$$
Since on has
$$\Large\mathop {\lim }\limits_{n \to \infty } \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \frac{1}{{{2^{n - 1}}}} = 1$$
and
$$\Large\mathop {\lim }\limits_{n \to \infty } \root n \of a = 1$$
for any real $a>0$, it is immeadiate that
$$\Large\lim {a_n} = {c^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \frac{1}{{{2^{n - 1}}}}}}{a^{\frac{1}{{{2^n}}}}} = {c^1} = c$$
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1This approach can easily be generalized, see my answer. – 2012-04-26
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0The statement about convergence of the nth root of $a$ is only valid for $a>0$. That condition holds here but isn't what you stated. – 2012-04-26
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1@Patrick The OP clearly stated $a$ is positive. Why the fuzz? It is rather trivial that for $a<0$ we will have trouble! – 2012-04-26
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0@Peter: Not a big deal, but it wasn't true as stated, that's all. I thought mathematicians were supposed to be picky :) – 2012-04-26
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0@Patrick I admit it said "for real $a$", but note the OP wrote $a_1>0$ and I gave the equivalence $a=a_1$, so I guess it wasn't too much to worry about. – 2012-04-26
From my answer here: Limit of sequence $x_n^n$
This is true because of the following Lemma:
Lemma: Suppose $\displaystyle z_n \gt 0$ is a sequence and $\displaystyle \alpha \gt 1$ is a real number such such that $$\lim_{n \to \infty} \frac{z_{n+1}^{\alpha}}{z_n} = q$$ then $$\lim_{n \to \infty} z_n = q^{1/(\alpha -1)}$$
Proof of Lemma
For the moment, assume that $\displaystyle q \gt 0$.
We have that, given an arbitrary $q \gt \varepsilon \gt 0$, there is some $n_0$ such that $\forall n \ge n_0$
$$ q - \varepsilon \lt \frac{z_{n+1}^\alpha}{z_n} \lt q+\varepsilon$$
$$ \sqrt[\alpha]{q- \varepsilon}\lt \frac{z_{n}}{\sqrt[\alpha]{z_{n-1}}} \lt \sqrt[\alpha]{q+\varepsilon}$$
$$ \dots $$
$$ \left(q - \varepsilon\right)^{1/\alpha^{n-n_0}} \lt \frac{z_{n_0+1}^{1/\alpha^{n-n_0 + 1}}}{z_{n_0}^{1/\alpha^{n-n_0}}} \lt\left(q + \varepsilon\right)^{1/\alpha^{n-n_0}} $$
Multiplying all and taking $\displaystyle \alpha^{th}$ root once gives us
$$C^{1/\alpha^{n-n_0+1}}\left(q - \varepsilon\right)^{1/(\alpha-1) - 1/\alpha^{n-n_0+1}}\lt z_{n+1} \lt C^{1/\alpha^{n-n_0+1}} \left(q + \varepsilon\right)^{1/(\alpha-1) - 1/\alpha^{n-n_0+1}}$$
Taking limits as $\displaystyle n \to \infty$ gives us
$$(q - \varepsilon)^{1/(\alpha-1)} \le \liminf z_{n} \le \limsup z_{n} \le (q+\varepsilon)^{1/(\alpha-1)}$$
Since $\displaystyle \varepsilon$ was arbitrary, we have that $\displaystyle \lim z_n = q^{1/(\alpha-1)}$.
If $\displaystyle q = 0$, all we need to do is replace the left hand side by $\displaystyle 0$ and the proof carries through.
Remark: The proof is similar to the textbook proof of $\lim \frac{a_{n+1}}{a_n} = \lim a_n^{1/n}$ which can be found in my answer here: https://math.stackexchange.com/a/116198/1102.
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0@PeterTamaroff: Yes, I did. And yes, the indices do make you dizzy :-) – 2012-04-26
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0I see the simpler "proof" rooting from $$\Large {a_n} = {c^{\sum\limits_{k = 1}^{n - 1} {\frac{1}{{{\alpha ^k}}}} }}{a^{\frac{1}{{{\alpha ^{n-1}}}}}}$$ but your approach is definitely more rigorous. – 2012-04-26
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0@PeterTamaroff: Yes, but that is only applicable to the sequence $(a_{n+1})^{\alpha} = c a_n$. Here, all we need is that the ratio have a limit. For instance, see the original problem (link to it in the first sentence) which prompted me to prove this. – 2012-04-26
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0Oh. You can see I really scanned through your answer. It is a greater generalization than I thought. Great one! I'll read it properly tomorrow. – 2012-04-26
I draw a picture with $c=5$,which will help you to understand the question.
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1OK, if it's a sequence of real numbers ($a_{n-1} = a_n^2/c \ge 0$). But if complex numbers are allowed, all you can say is $a_n = \pm \sqrt{c a_{n-1}}$, and there won't necessarily be a limit. – 2012-04-26
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0@RobertIsrael: $a_1 \gt 0$ and $c \gt 0$, so I presume we are only dealing with real numbers... – 2012-04-26
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0@Aryabhata: Not necessarily. $a_2$ could be negative and then $a_3$ is imaginary. – 2012-04-26
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0@RobertIsrael: You are right, but I was talking about the problem likely making an implicit assumption that they are reals (which is usually the case for basic problems like these). If we allow $a_i$ to be complex, I don't see the point of stating $c \gt 0$ and $a_1 \gt 0$. – 2012-04-26
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0@Aryabhata: I agree that the author of the problem probably intended to imply $a_k$ real, but did not succeed in doing that. – 2012-04-26
If the sequence converges, it certainly converges to either $c$ or 0, since $$\lim_{n\to\infty}(a_n)^2=\lim_{n\to\infty}ca_{n-1}$$ $$(\lim_{n\to\infty}a_n)^2=c\lim_{n\to\infty}a_{n-1}=c\lim_{n\to\infty}a_{n}$$
Now, we need to show this sequence is Cauchy, and that the limit cannot be 0. I claim the sequence is bounded and monotone. If $a_1>c$, then by induction $a_{n-1}>c$, so $a_n^2=ca_{n-1}>c^2$, hence $a_n>c$, but $a_n=\sqrt{ca_{n-1}}<\sqrt{a_{n-1}^2}=a_{n-1}$, hence the sequence is decreasing and bounded from below by $c$. The sequence is thus bounded and monotone, so the limit exists, and we have shown by induction that $a_1>a_n>c>0$, so $a_1\geq\lim_{n \to \infty}a_n\geq c>0$, hence the limit is not 0.
The proof for when $a_1<c$ is essentially the same.
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0From the second displayed line, the limit could be $0$ as well, whereas $c>0$. That it's not requires at least some argument. – 2012-04-26
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0Good point. If you at the induction part of my post, though, I prove by induction that $a_1 \geq\lim a_n\geq c$, which takes care of that issue. – 2012-04-26
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1Right, but in your first line you claim that convergence implies that the limit is $c$. In this case that is the limit, but the correct statement in the first line would be that it certainly converges to either $c$ or $0$. – 2012-04-26
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1@Patrick I like your nitpickyness! – 2012-04-26
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0@Patrick Alright, I've edited it directly into the post instead of just leaving this discussion in the comments. – 2012-04-26