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?
 A: 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
$$\Large\lim {a_n} = {c^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} +  \cdots \frac{1}{{{2^{n - 1}}}}}}{a^{\frac{1}{{{2^n}}}}} = {c^1} = c$$
A: 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.

A: I draw a picture with $c=5$,which will help you to understand the question.
A: 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 $a_n=\sqrt{ca_{n-1}}&lt\sqrt{a_{n-1}^2}=a_{n-1}$, 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 $a_1&ltc$ is essentially the same.
A: 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 $e^x$. 
