Convergence of a sequence defined by $ c_{n+1}=\frac{r_{n+1}\sqrt{c_n}}{3}$ 
Define a sequence $(r_n)$ by $r_0=1$ and $r_{n+1}=(2/3)r_n+1$ for $n\geq 0$. Let the sequence $(c_n)$ be defined by $c_0=1/4$, and
  $$
c_{n+1}=\frac{r_{n+1}\sqrt{c_n}}{3}
$$
  for $n\geq 0$. Prove that $\lim_{n\to\infty}c_n$ exists. 


It is not difficult to prove that the sequence $(r_n)$ converges to $3$ since it is bounded above and non-decreasing. I don't have a more useful estimate for $(c_n)$ than $c_{n+1}\leq \sqrt{c_n}$. How shall I go on?
 A: Enlightened by answers and comments, I have a proof. 
One can see that $r_{n+1}-3=\frac{2}{3}(r_n-3),$ which implies by induction that 
$$
r_{n}-3=(r_0-3)q^n,\quad q=2/3,
$$
and thus 
$$
c_{n+1}=\left(1-\left(\frac{2}{3}\right)^{n+2}\right)\cdot\sqrt{c_n}=:b_n\sqrt{c_n}. 
$$
On the other hand, $c_1\geq c_0$ by direct calculation. Suppose $c_{n+1}\geq c_n$. Then 
$$
c_{n+2}=b_{n+1}\sqrt{c_{n+1}}\geq b_n\sqrt{c_{n+1}}\geq b_n\sqrt{c_n}=c_{n+1}.
$$
It follows that $(c_n)$ is a non-decreasing sequence. 
All we need to do now is showing that $(c_n)$ is bounded above. We use induction again. Obviously $c_0\leq 1$. Suppose $c_n\leq 1$. Then 
$$
c_{n+1}\leq\sqrt{c_n}\leq 1. 
$$
We are done. 
A: Putting the recursion of $\,r_n\,$ in the recursion of $\,c_n\,$ we get 
$\displaystyle b_n:=\frac{c_{n+1}}{\sqrt{c_n}}= \frac{2}{3} b_{n-1}+\frac{1}{3}=…=1+(\frac{2}{3})^n(b_0-1)$
with $\enspace\displaystyle b_0= \frac{c_1}{\sqrt{c_0}} =\frac{r_1}{3} =\frac{1}{3} (\frac{2}{3}r_0 +1)= \frac{5}{9} $ .
The recursion for $\,c_n\,$ is now $\enspace\displaystyle c_n=b_{n-1} \sqrt{c_{n-1}}=(1-\left(\frac{2}{3}\right)^{n+1})\sqrt{c_{n-1}} $ .
It's $\enspace\displaystyle \ln c_n=-\frac{\ln 2}{2^{n-1}}+\sum\limits_{k=0}^{n-1}\frac{1}{2^k}\ln\left(1-\left(\frac{2}{3}\right)^{n+1-k}\right)$
and with $\enspace\displaystyle \ln\frac{5}{9}\leq \ln\left(1-\left(\frac{2}{3}\right)^{n+1-k}\right)<0$  
we get $\enspace\displaystyle -\frac{\ln 2}{2^{n-1}}+2(\ln\frac{5}{9})(1-\frac{1}{2^n})\leq\ln c_n<-\frac{\ln 2}{2^{n-1}}$
and therefore $\enspace\displaystyle \frac{25}{81}\leq c_\infty \leq 1$ .
$\enspace\displaystyle c_0=\frac{1}{4}<\frac{25}{81}\enspace$ implies that the sequence of $\,(c_n)\,$ is increasing 
and therefore exists a well defined limit $\,c_\infty\,$ .  
Now we get directly $\,c_\infty=1\,$ from the recursion for $\,c_n\,$ above. 
Note:
The easiest way to show directly that the sequence $(c_n)$ is increasing is induction, which is mentioned in Jack's answer.   
