Prove that the sequence$ c_1 = 1$, $c_{n+1} = 4/(1 + 5c_n) $ , $ n \geq 1$ is convergent and find its limit 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.
 A: Here is a more explicit version of my hint.  I said it would be easier to show that $d_n = c_n - \frac{4}{5}$ tends to zero as $n \to \infty$.  This gives $d_1 = \frac{1}{5}$ and the recurrence
$$d_{n+1} + \frac{4}{5} = \frac{4}{1 + 5(d_n + \frac{4}{5})} = \frac{4}{5} \frac{1}{1 + d_n}$$
hence
$$d_{n+1} = - \frac{4}{5} \frac{d_n}{1 + d_n}.$$
Can you see what to do from here?  
Edit:  Some more hints.  If the $1 + d_n$ in the denominator were just a $1$, we would be done because then $|d_n|$ would decrease exponentially at least as fast as $\left( \frac{4}{5} \right)^n$.  But it's not.  However, $d_n$ is small, so the denominator should be close enough to $1$ that this argument should still go through.  More precisely, you just need to find a constant $0 < c < \frac{1}{5}$ such that you can prove that $|d_n| \le c$, say for $n \ge 2$.  (Maybe by induction.)  From there it will follow that $|d_n|$ actually decays exponentially at least as fast as $\left( \frac{4}{5(1 - c)} \right)^n$.
This is an example of a general technique called "bootstrapping," where you use weak estimates together with relations that a sequence satisfies to get stronger estimates.
A: Well done! Your observations are correct and can be completed to give a proof that the sequence is convergent to $\frac{4}{5}$.
The tough part in proving your observations it to prove that the odd/even sub-sequences are monotonic and appropriately bounded (what do I mean by this?).
Here is a hint:
Show that $$ c_{2n-1} \ge \frac{4}{5} \ge c_{2n} \ \ \forall n \ge 1$$
Hint2:
Try writing $\displaystyle c_{n+2}$ in terms of $\displaystyle c_{n}$ and see if that helps you prove the above bounds (and as a next step, the monotonicity).
(For a simpler proof of the above bound, you can also try showing that if $c_{n} \ge 4/5$ then $c_{n+1} \le 4/5$ and similarly if $c_n \le 4/5$ then $c_{n+1} \ge 4/5$).
A: 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.
A: Here's one way to prove it: let $f(x) = 4/(1+5x)$. Say $|x-4/5| \le C$ for some constant $C$. Can you find $C$ and some constant $0 \le k < 1$ so that if $|x-4/5| \le C$, then $|f(x)-4/5| \le k|x-4/5|$? 
If you do this, then you can iterate to get $|f^j(x)-4/5| \le k^j |x-4/5|$, for all $j$, and so if you make $j$ large enough then you can get $f^j(x)$ as close to $4/5$ as you like.
A: So $c_{1} = 1$ and $c_{n+1} = 4/(1+5c_{n})$ for $n \geq 1$. Let $C(x) = \sum_{n \geq 1} c_{n}x^n$. Then maybe try to express $\sum_{n \geq 1} c_{n+1}x^n$ and $\sum_{n \geq 1} \frac{4x^n}{1+5c_{n}}$ in terms of $C(x)$ to get a closed form. 
