Question: I was trying to use generating functions to get a closed form for the sum of first $n$ positive integers.

So far I got to $G(x) = \frac{1}{(1-x)^3}$ but I don't know how to convert this back to the coefficient on power series.

Answer: 
You can use $\frac 1{1-x}=\sum\_{i=0}^\infty x^i$ and take two derivatives
$$\frac {d^2}{dx^2}\frac 1{1-x}=\frac 2{(1-x)^3}=\frac {d^2}{dx^2}\sum\_{i=0}^\infty x^i=\sum\_{i=0}^\infty(i+2)(i+1)x^i\\=1+3x+6x^2+10x^3+\ldots$$
which is off by $1$ in the index.
Since $\frac{1}{1-x}=1+x+x^2+x^3+\ldots$ for any $x\in(-1,1)$, we have

$$ [x^n]\frac{1}{(1-x)^k} = \begin{array}{c}\small\text{number of ways for writing }n\\\small\text{as a sum of }k\small\text{ natural numbers}\end{array} $$
and the RHS is given by [stars and bars](https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)). In general we have
$$ \frac{1}{(1+x)^{m+1}} = \sum\_{n\geq 0}\binom{m+n}{n}x^n.$$