closed form of a Cauchy (series) product I hope this hasn't been asked already, though I have looked around the site and found many similar answers.
Given:
Form Cauchy product of two series: $a_k\;x^k$ and $\tfrac{1}{1-x}=1+x+x^2+\cdots$. 

So I come up with,
$\sum_{n=0}^{\infty}\;c_n = \sum_{n=0}^{\infty}\;\sum_{k+l=n}\;a_l\;b_k = \cdots = \sum_{n=0}^{\infty}\;x^n\;\sum_{k=0}^n\;a_k = x^0\;(a_0)+x^1\;(a_0+a_1)+x^2\;(a_0+a_1+a_2)+\cdots$.
It asks for what values of $x$ this would be valid:
This is a funny question to me because it depends upon the coefficients in the power series, right? If I take the ratio test, I get $\lim_{n\to\infty}\;\bigg| \frac{a_{k+1}\;x^{k+1}}{a_k\;x^k}\bigg| = |x|\cdot \lim_{k\to\infty}\;\big| \frac{a_{k+1}}{a_k} \big|$. For this series to be convergent, doesn't this have to come to a real number, $L$ (not in $\mathbb{\bar{R}}$)? Therefore, $|x|<1/L$?

I know that the other series, $\sum_{n=0}^{\infty}\;x^n$ converges for $r \in (-1,1)$. 

So for the product to be convergent, doesn't the requirement of $x$ have to be, $|x| < \min\{1,1/L\}$?

The reason I include this, other than the questions above, is that the question suggests using "this approach" to attain a closed form $\sum_{k=0}^{\infty}\;k\;x^k$, for $x \in (-1,1)$. By using the ratio test (my favorite), I'm pretty sure that for this to converge, $|x|<1$ - which is given. I tried writing out some of the terms but they do not seem to reach a point whereby future terms cancel (as they do in a series like $\sum_{n=1}^{\infty} \frac{1}{k\;(k+2)\;(k+4)}$). I've tried bounding (Squeeze) them but didn't get very far.
Thanks for any suggestions!

Note that $\sum_{n\ge 0}nx^n$ is almost the Cauchy product of $\sum_{n\ge 0}x^n$ with itself: that Cauchy product is 
$$\left(\sum_{n\ge 0}x^n\right)^2=\sum_{n\ge 0}x^n\sum_{k=0}^n 1^2=\sum_{n\ge 0}(n+1)x^n\;.\tag{1}$$
If you multiply the Cauchy product in $(1)$ by $x$, you get
$$x\sum_{n\ge 0}(n+1)x^n=\sum_{n\ge 0}(n+1)x^{n+1}=\sum_{n\ge 1}nx^n=\sum_{n\ge 0}nx^n\;,\tag{2}$$
since the $n=0$ term is $0$ anyway. Combining $(1)$ and $(2)$, we have
$$\sum_{n\ge 0}nx^n=x\left(\sum_{n\ge 0}x^n\right)^2=x\left(\frac1{1-x}\right)^2=\frac{x}{(1-x)^2}\;,$$
with convergence for $|x|<1$ by the reasoning that you gave.
