Coefficients in series expansion of $\left(\frac{x}{1-x}\right)^3$ How can I find the coefficient of $x^m$ in 
$$
\left(\sum_{n=1}^{\infty}x^n\right)^3
$$
for some $m\in{\Bbb N}$?
I attempted it using the Cauchy products, but things got messy. I think one might want to write it as
$$
\left(\frac{x}{1-x}\right)^3
$$
But I have no idea how to find the Taylor series of this function.
 A: Rewrite this as $x^3(1-x)^{-3}$, write the taylor expansion of $(1-x)^{-3}$, then multiply it by $x^3$
A: Put $x^3(1-x)^{-3} = \sum_{m \ge 0} a_m x^m$, and multiply by $(1-x)^3 = 1-3x+3x^2-x^3$ : 
you get $x^3 = a_0 + (a_1-3a_0)x^1 + (a_2-3a_1+3_a0)x^2 + \sum_{m \ge 3} (a_m-3a_{m-1}+3a_{m-2}-a_{m-3}) x^m$
, from which you deduce that $a_0 =a_1 = a_2 = 0, a_3 = 1$, and forall $m\ge4$, $a_m = 3a_{m-1}-3a_{m-2}+a_{m-3}$.
This last statement is equivalent to saying that the function $m \mapsto a_m$ is a polynomial in $m$ of degree $2$ on $\{1,2,3,\ldots\}$.
Since $a_1 = a_2 = 0$ and $a_3 = 1$ we must have $a_m = \frac {(m-1)(m-2)}{(3-1)(3-2)} = \frac {(m-1)(m-2)}2$ forall $m \ge 1$.
Alternatively, you can just develop the product, observe that $a_m$ is the number of integer triples $(i,j,k)$ such that $i,j,k \ge 1$ and $i+j+k = m$, and try to count them. 
A: The usual power series expansion $(1-x)^{-3} = \sum_{k\geq 0} \left( \begin{matrix} -3 \\ k \end{matrix} \right) (-x)^3$ works (in fact for any complex number instead of $-3$). Here, 
$$ \left( \begin{matrix} -3 \\ k \end{matrix} \right) = \frac{(-3)(-4) \cdots (-3-k+1)}{k!}.$$ Proof by Taylor expansion (as already mentioned).
