Taylor Series Expansion of $f(x) = 2/(1-x)$ centered at x=3. Give the interval of convergence.

Taylor Series Expansion of $f(x) = 2/(1-x)$ centered at x=3. Give the
interval of convergence.

Find the Taylor Expansion for the function f(x) = 2/(1-x) centered at x =
3. Give the interval of convergence for this series.
So if I remember correctly, we first take the first four or so derivatives.
$f(x) = 2/(1-x)$
$f'(x) = 2/(1-x)^2$
$f''(x) = 4/(1-x)^3$
$f'''(x) = 12/(1-x)^4$
$f''''(x) = 48/(1-x)^5$



Now at x = 3 for those derivatives are:
$f(3) = -1$
$f'(3) = .5$
$f''(3) = -.5$
$f'''(3) = (3/4)$
$f''''(3) = -(3/2)$



Is this the Taylor Expansion?
$f(x) = -1 + (1/2)(x-3) - (1/2)( (x-3)^{2} / 2! ) + (3/4)( (x-3)^{3} / 3!
) ...$
I need to find a general solution for this, but I can't really nail down a
pattern. To do the interval of convergence portion I need to figure out
that pattern so this is where I am stuck.