Finding a closed formula for a generating function involving catalan numbers For $n$ $\ge$ $1$, let the Catalan number $C_n$ be defined to be the number of ways of partitioning a convex $(n+2)$-gon into $n$ triangles by using diagonals that do not cross one another (except perhaps at their ends). By convention set $C_0$ $=$ $1$. 
Find a recurrence relation expressing $C_n$ in terms of the Catalan numbers of smaller index.
I have found this to be $C_{n+1}$ $=$ $\frac{4n+2}{n+2}C_n.$
I want to use this to find a closed formula for the generating function $$g(x) = \sum_0^{\infty}C_nx^n.$$
Please help me..

Going directly from your recursion is not to difficult, but first you need to correct the recursion -- your index is off by 1. The right relation is
$$
C_k = C_{k-1} \frac{4k+2}{k+2} + [k = 0]
$$
where $[k = 0]$ stands for a function of $k$ which is $1$ when $k=0$ and zero otherwise.
We define (for this derivation) $C_{n} = 0$ for all negative $n$.
Break up
$$ \frac{4k+2}{k+2} = 4 -  \frac{8k+8 -6}{k+2} =4-\frac{6}{k+2}$$
to get (R)
$$
C_k = 4 C_{k-1} -6 \frac{C_{k-1}}{k+2} + [k = 0]
$$
Multiply (R) by $z^k$ for all values of $k$ and sum those equations to get a generating function relation:
$$
\sum_{k=0}^\infty C_k z^k= 4 \sum_{k=0}^\infty C_{k-1}z^{k} -6\sum_{k=0}^\infty \frac{1}{k+2}C_{k-1}z^{k}+1
$$
Here, we have used the fact that $[k=0]$ contributes 1 to the sum when $k$ is zero, and nothing else.
Now shift the summation indices on the right by using $m=k-1$ so that only $C_d$ appears, where $d$ is a "dummy" (summed over) index, to get (S):
$$
C(z) \equiv \sum_{k=0}^\infty C_k z^k= 4 \sum_{m=0}^\infty C_{m}z^{m+1} -6\sum_{m=0}^\infty \frac{1}{m+1+2}C_{k-1}z^{m+1}+1
\\=4 z C(z) - 6\sum_{m=0}^\infty \frac{1}{m+3}C_{k-1}z^{m+1}+1
$$
The $(m+3)$ in the denominator would disappear if we had a $z^{m+3}$ multiplying it, and differentiated that term.  So multiply (S) by $z^2$ and differentiate with respect to $z$:
$$\frac{d}{dz} \left( z^2 C(z) \right) = 
4 \frac{d}{dz} \left( z^3 C(z) \right) + z^2 C(z) + \frac{d}{dz}(z^2)
$$
Use the multiplication rule for derivatives to simplify this and get (T):
$$
z^2 \frac{dC(z)}{dz} + 2z C(z) = 4z^3 \frac{dC(z)}{dz} +12 z^2 C(z) - 6 z^2 C(z) + 2 z\\
(4z^3-z^2)  \frac{dC(z)}{dz} + (6 z^2 -2 z) C(z)  + 2 z = 0
$$
Solve differential equation (T) to get
$$
C(z) = \frac{1-2z+\kappa \sqrt{1-4z}}{2z^2}
$$
The behavior of this function near $z=0$ is
$$
\frac{1+\kappa}{2z^2} - \frac{1+\kappa}{z} -\kappa + \cdots
$$
and since the coefficients for negative exponents of $z$ need to be zero, this determines that $\kappa = -1$.
$\kappa$ can be determined by noting that the constant term of the generating function must be $C_0 = 1$, and the which forces $\kappa = 1$. 
So the generating function for $C_k$ is
$$
C(z) = \frac{1-2z-\sqrt{1-4z}}{2z^2}
$$
