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.
Let $\frac{1}{(1-x)^3}=\sum\_{i=0}^{\infty}a\_ix^i$, then $$1=(1-3x+3x^2-x^3)\sum\_{i=0}^{\infty}a\_ix^i=\sum\_{i=0}^{\infty}a\_ix^i-\sum\_{i=0}^{\infty}3a\_ix^{i+1}+\sum\_{i=0}^{\infty}3a\_ix^{i+2}-\sum\_{i=0}^{\infty}a\_ix^{i+3}$$

From this we get (by comparing LHS and RHS):

* $1=a\_0$
* $0=a\_1-3a\_0\Rightarrow a\_1=3$
* $0=a\_2-3a\_1+3a\_0\Rightarrow a\_2=6$
* $0=a\_{n+3}-3a\_{n+2}+3a\_{n+1}-a\_n$, for $n\geq 0$

Now, consider the difference between each coefficient $$a\_0,a\_1,a\_2,a\_3,a\_4,a\_5,\dots\\a\_1-a\_0,a\_2-a\_1,a\_3-a\_2,a\_4-a\_3,a\_5-a\_4,\dots\\a\_2-2a\_1+a\_0,a\_3-2a\_2+a\_1,a\_4-2a\_3+a\_2,a\_5-2a\_4+a\_3,\dots\\0,0,0,0,0,\dots$$ 

The last line is $0$, since $0=a\_{n+3}-3a\_{n+2}+3a\_{n+1}-a\_n$. This means that $a\_n$ is a polynomial of degree $2$, so that $$a\_n=c\_2n^2+c\_1n+c\_0$$

To find $c\_2,c\_1,c\_0$ note:

* $a\_0=1\Rightarrow c\_0=1$
* $a\_1=3\Rightarrow 2=c\_2+c\_1$
* $a\_2=6\Rightarrow5=4c\_2+2c\_1$

This gives:

* $c\_0=1$
* $c\_1=\frac{3}{2}$
* $c\_2=\frac{1}{2}$

So: $$a\_n=\frac{n^2}{2}+\frac{3n}{2}+1=\frac{(n+1)(n+2)}{2}$$