Originally Posted by

**angel.white** Find a power series representation for the function and determine the radius of convergence:

$\displaystyle f(x) = \frac {x^3}{(x-2)^2}$

This is what I did:

$\displaystyle = x^3 \left( \frac {1}{x-2} \right)^2$

$\displaystyle = x^3 \left( -\frac12 * \frac {1}{1-\frac x2}\right)^2$

$\displaystyle = x^3 \left( -\frac12 * \sum_{n=0}^\infty \left( \frac x2 \right)^n\right)^2$

So radius = 2

But I don't know where to go from here, the answer says:

$\displaystyle \sum_{n=3}^\infty \frac {n-2}{2^{n-1}} x^n$