Use Cauchy product to find a power series represenitation of $1 \over {(1-x)^3}$ 
Use Cauchy product to find a power series represenitation of
  $$1 \over {(1-x)^3}$$ which is valid in the interval $(-1,1)$.

Is it right to use the product of $1 \over {1-x}$ and $1 \over {(1-x)^2}$ if I know thier expantion (HOW).. Is there another way?
 A: This is most easily done with the binomial theorem and the identity
$$
\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}
$$
which gives
$$
\begin{align}
\frac1{(1-x)^3}
&=\sum_{k=0}^\infty\binom{-3}{k}(-x)^k\\
&=\sum_{k=0}^\infty\binom{k+2}{k}x^k\\
&=\sum_{k=0}^\infty\frac{(k+2)(k+1)}{2}x^k
\end{align}
$$

To use the Cauchy products, we would use the $n=1$ version of the series above
$$
\frac1{1-x}=\sum_{k=0}^\infty x^k\tag{1}
$$
The Cauchy product method says that
$$
\left(\sum_{k=0}^\infty a_k x^k\right)\left(\sum_{k=0}^\infty b_k x^k\right)
=\sum_{k=0}^\infty\left(\sum_{j=0}^k a_jb_{k-j}\right)x^k\tag{2}
$$
$(2)$ applied to $(1)$ twice gives
$$
\begin{align}
\frac1{(1-x)^2}
&=\sum_{k=0}^\infty\left(\sum_{j=0}^k1\cdot1\right)x^k\\
&=\sum_{k=0}^\infty(k+1)x^k\tag{3}
\end{align}
$$
$(2)$ appplied to $(1)$ and $(3)$ yields
$$
\begin{align}
\frac1{(1-x)^3}
&=\sum_{k=0}^\infty\left(\sum_{j=0}^k(j+1)\cdot1\right)x^k\\
&=\sum_{k=0}^\infty\frac{(k+2)(k+1)}{2}x^k\tag{4}
\end{align}
$$
A: It is faster with term-by-term differentiation, but since you asked for a Cauchy product argument, here is one.
The series 
$$
\sum_{n\geq 0}x^n=\frac{1}{1-x}
$$
converges absolutely for every $x$ in $(-1,1)$. From now on, we fix $x\in (-1,1)$.
So its Cauchy product with itself converges absolutely and yields
$$
\left(\frac{1}{1-x}\right)^2=\left(\sum_{n\geq 0}x^n\right)^2=\sum_{n\geq 0}\sum_{i+j=n}x^ix^j=\sum_{n\geq 0}x^n\sum_{i+j=n}1=\sum_{n\geq 0}(n+1)x^n.
$$
Indeed, the general term $c_n$ of the Cauchy product of $\sum_{n\geq 0} a_n$ by $\sum_{n\geq 0} b_n$ is 
$$c_n=\sum_{i+j=n}a_ib_j=\sum_{i=0}^na_ib_{n-i}=\sum_{j=0}^na_{n-j}b_{j}.$$
In this case, we find $x^n\sum_{j=0}^n1=(n+1)x^n$.
Do one more Cauchy product with the initial series to get
$$
\left(\frac{1}{1-x}\right)^3=\left(\sum_{n\geq 0}x^n\right)\left(\sum_{n\geq 0}(n+1)x^n\right)=\sum_{n\geq 0}\sum_{i+j=n}x^i(j+1)x^j=\sum_{n\geq 0}x^n\sum_{i+j=n}j+1.
$$
So the general term of the Cauchy product is $$x^n\sum_{j=0}^n(j+1)=x^n(1+2+\ldots+(n+1))=\frac{(n+1)(n+2)}{2}x^n.$$ 
Hence

$$
\left(\frac{1}{1-x}\right)^3=\sum_{n\geq 0}\frac{(n+1)(n+2)}{2}x^n\qquad \forall x\in (-1,1).
$$

A: This can be done using Cauchy product formula for a product of three series. We have $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$, and hence
$$
\frac{1}{(1-x)^3} = \sum_{n=0}^\infty a_n x^n,
$$
where
$$
a_n = \sum_{i,j,k\geq 0, i+j+k=n} 1 \cdot 1 \cdot 1 = \binom{n+2}{n} = \frac{(n+1)(n+2)}{2}.
$$
A: Recall the geometric series: 
$$
\sum_{j=0}^\infty x^{j}=\frac{1}{1-x}\quad \text{if }x\in (-1,1)
$$
Then, you can use the Cauchy product twice, as 
$$
\frac{1}{(1-x)^3}=\frac{1}{1-x}\frac{1}{1-x}\frac{1}{1-x}
$$
This isn't the easiest way to do it, but if you're required to use the Cauchy product formula, this is how you would do it.
A: It is worth noting that $\frac{d^2}{dx^2}[\frac{1}{1-x}]=\frac{2}{(1-x)^3}$. Thus $\frac{1}{(1-x)^3}=\frac{1}{2}\frac{d^2}{dx^2}[\frac{1}{1-x}]=\frac{1}{2}\frac{d^2}{dx^2}\Sigma_{i=0}^{\infty}x^i=\Sigma_{i=0}^{\infty}\frac{i(i-1)}{2}x^{i-2}=\Sigma_{i=2}^{\infty}\frac{i(i-1)}{2}x^{i-2}=\Sigma_{i=0}^{\infty}\frac{(i+2)(i+1)}{2}x^{i}$.
The point of this is that sometimes we can compute these products in different ways. In this instance, we were able to differentiate something we knew to get the desired answer. 
