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.
Using [Binomial series](https://en.wikipedia.org/wiki/Binomial_series),

for $|-x|<1,$

$$(1-x)^{-3}=\sum\_{r=0}^\infty\dfrac{(-3)(-2)\cdots\{-3-(r-1)\}}{r!}(-x)^r$$

Now $\dfrac{(-3)(-3-1)\cdots\{-3-(r-1)\}}{r!}=(-1)^r\dfrac{(r+2)(r+1)\cdots3}{r!}=(-1)^r\dfrac{(r+1)(r+2)}2$