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 $a_n$?
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.
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 $L=+2$.