# 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 $c$.

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?

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Although my mother language is Spanish (too?) I guess it is necessary to post in English here. I "transledited". –  Pedro Tamaroff Apr 26 '12 at 0:29

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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From the second displayed line, the limit could be $0$ as well, whereas $c>0$. That it's not requires at least some argument. –  Patrick Apr 26 '12 at 0:55
Good 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. –  Brett Frankel Apr 26 '12 at 0:56
Right, 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 $0$. –  Patrick Apr 26 '12 at 1:09
@Patrick I like your nitpickyness! –  Pedro Tamaroff Apr 26 '12 at 1:11
@Patrick Alright, I've edited it directly into the post instead of just leaving this discussion in the comments. –  Brett Frankel Apr 26 '12 at 16:20

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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This approach can easily be generalized, see my answer. –  Aryabhata Apr 26 '12 at 1:00
The 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. –  Patrick Apr 26 '12 at 1:06
@Patrick The OP clearly stated $a$ is positive. Why the fuzz? It is rather trivial that for $a<0$ we will have trouble! –  Pedro Tamaroff Apr 26 '12 at 1:08
@Peter: Not a big deal, but it wasn't true as stated, that's all. I thought mathematicians were supposed to be picky :) –  Patrick Apr 26 '12 at 1:19
@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. –  Pedro Tamaroff Apr 26 '12 at 1:47

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: http://math.stackexchange.com/a/116198/1102.

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I really like generalizations! However, it is kinda late here and all those indices are getting me dizzy! (+1, I guess you also upvoted my answer.) –  Pedro Tamaroff Apr 26 '12 at 1:02
@PeterTamaroff: Yes, I did. And yes, the indices do make you dizzy :-) –  Aryabhata Apr 26 '12 at 1:03
I 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. –  Pedro Tamaroff Apr 26 '12 at 1:06
@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. –  Aryabhata Apr 26 '12 at 1:08
Oh. 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. –  Pedro Tamaroff Apr 26 '12 at 1:10

I draw a picture with $c=5$,which will help you to understand the question.

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OK, 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. –  Robert Israel Apr 26 '12 at 0:56
@RobertIsrael: $a_1 \gt 0$ and $c \gt 0$, so I presume we are only dealing with real numbers... –  Aryabhata Apr 26 '12 at 1:15
@Aryabhata: Not necessarily. $a_2$ could be negative and then $a_3$ is imaginary. –  Robert Israel Apr 26 '12 at 6:56
@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 $a_1 \gt 0$. –  Aryabhata Apr 26 '12 at 6:58
@Aryabhata: I agree that the author of the problem probably intended to imply $a_k$ real, but did not succeed in doing that. –  Robert Israel Apr 26 '12 at 7:23
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 $e^x$.