TITLE: Prove that the sequence$ c_1 = 1$, $c_{n+1} = 4/(1 + 5c_n) $ , $ n \geq 1$ is convergent and find its limit
QUESTION [11 upvotes]: Prove that the sequence $c_{1} = 1$, $c_{(n+1)}= 4/(1 + 5c_{n})$     , $n \geq 1$  is convergent and find its limit.
Ok so up to now I've worked out a couple of things.
$c_1 = 1$
$c_2 = 2/3$
$c_3 = 12/13$
$c_4 = 52/73$  
So the odd $c_n$ are decreasing and the even $c_n$ are increasing.  Intuitively, it's clear the the two sequences for odd and even $c_n$ are decreasing/increasing less and less.
Therefore it seems like the sequence may converge to some limit $L$.
If the sequence has a limit, let $L=\underset{n\rightarrow \infty }{\lim }a_{n}.$   Then $L = 1/(1+5L).$
So we yield $L = 4/5$ and $L = -1$.  But since the even sequence is increasing and >0, then $L$ must be $4/5$.
Ok, here I am stuck.  I'm not sure how to go ahead and show that the sequence converges to this limit (I tried using the definition of the limit but I didn't manage) and and not sure about the separate sequences how I would go about showing their limits.
A few notes :
I am in 2nd year calculus.
This is a bonus question, but I enjoy the challenge and would love the extra marks.
Note : Once again I apologize I don't know how to use the HTML code to make it nice.

REPLY [2 votes]: As shown by the other answers, there are a few nice ways to approach this problem.
You could concentrate only on $C_{2n-1},$ say, since if you establish that $C_{2n-1}$ tends to a limit you automatically nail $C_{2n}$ as well, because
$$C_{2n} = \frac{4}{1+5C_{2n-1}} \textrm { so }  \lim C_{2n}  = \lim  \frac{4}{1+5C_{2n-1}}.$$
Now it's easy to show $C_{2n-1} > 4/5$ and so expand
$$(5C_{2n-1}-4)(C_{2n-1}+1)>0$$
and manipulate (add $20C_{2n-1}$ to both sides and take the 4 to the RHS) to obtain
$$ C_{2n-1} > \frac{4+20C_{2n-1}}{21+5C_{2n-1}} = C_{2n+1},$$
and since $C_{2n-1}$ is bounded below by 4/5 the result follows immediately.