Proving the convergnce of a sequence So, I have to prove that the sequence defined as 
$a_{n+1}=\frac{6(1+a_n)}{7+a_n}$ converges and then find the limit.
I have few questions; Do i have to assume that $a_n \geq 0$ or $a_n \leq0$. Because if i assume that $a_n \geq 0$, I can prove that 
$\left|a_{n+1}-2\right| \leq \frac{4}{7}\left|a_{n}-2\right|$, and by induction, it would be easy to show that $a_n \to 2$.
Any help would be appreciated. Is there a way to prove convergence without knowing the signs of $a_n$?
 A: If you want an easier way to find what value the limit converges to, consider that by the definition of convergence, as $n\to\infty$, $a_{n+1} = a_n$.
Thus, let's let $a_{n+1} = a_n = L$.
Substituting this in, we get:
$$a_{n+1}=\frac{6(1+a_n)}{7+a_n}$$
$$L = \frac{6(1+L)}{7+L}$$
$$L = \frac{6 + 6L}{7+L}$$
$$7L + L^2 = 6 + 6L$$
$$L ^2 + L - 6 = 0$$
$$(L+3)(L-2) = 0$$
Therefore, $L = -3, 2$
Since all of the terms in this sequence are positive, $L = -3$ s an extraneous solution.
Therefore, the sequence converges to $2$.
As for proving that the sequence converges, assume that $a_n \ge 0$ and use induction. 
A: Convergence: Note that $$a_{n+1}=6-\frac{36}{7+a_n}$$ 
The function $\displaystyle f(x) = 6 - \frac{36}{7+x}$ has derivative $\displaystyle f'(x) =  \frac{36}{(7+x)^2}.$ So we have $0 \leq f'(x) <1$ so the sequence converges by the Contraction Mapping Theorem.
A: You definitely have to make some assumption, because the sequence does not converge in all cases.  For example, if you specify $a_0=-463/133$ as a starting value, then (if my arithmetic is correct) $a_1=-55/13$, $a_2=-7$, and $a_3$ is undefined.  
It's easy to show that there is an infinite sequence of such "bad" starting values, and it's possible to show that that sequence converges to $-3$ (which is the "extraneous solution" in Varun Iyer's answer).  Any statement that the original sequence converges must, at the very least, include a condition that rules out these "bad" values.  The simplest such condition is to say $a_0\ge0$, which clearly implies $a_n\ge0$ for all $n$.
A: As Varun showed, the fixed points obeying $L=6(1+L)/(7+L)$ are $+2$ and $-3$. For $a_0\gt -3$, one may see that one is getting close to $L=+2$. For $a_0\gt -3$, the subsequent terms are getting more negative, but they inevitably drop below $a_n\lt -7$ at some point, and for $a_0\lt -7$, $a_1$ may already be calculated to be positive i.e. $a_1\gt - 3$, because the graph of $y=6(1+x)/(7+x)$ is a hyperbola, and one starts to converge towards the limit $L=2$, anyway.
As Arpit wrote, Contraction Mapping Theorem guarantees the existence of a limit. The fixed points $L=+2$ and $L=-3$ are the only possible limits. And the sequence diverges away from $L=-3$ for almost all $a_0$. The only exception occurs when $a_0=-3$ when all $a_n=-3$ and $L=-3$ is also the limit. For all other values of $a_0$, the limit has to be $L=+2$.
