Exploiting geometric series with power series to find Taylors series

I'm confused between some formulae so I'm going to give some examples and you can let me know if what I'm writing is correct.

**Find the Taylor series for....**

**EXAMPLE 1:**

$\displaystyle f(x) = \frac{1}{1- (x)}$ around $\displaystyle x = 2$

Then,

$\displaystyle \frac{1}{1-(x)} = \frac{1}{3-(x+2)} = \frac{1}{3} \left( \frac{1}{1 - \frac{(x+2)}{3}} \right) = \frac{1}{3} \sum_{n=0}^{\infty} (\frac{x+2}{3})^{n} = \sum_{n=0}^{\infty} \frac{(x+2)^{n}}{3^{n+1}}$ provided that $\displaystyle | \frac{x+2}{3} | < 1$ or, $\displaystyle -5 < x < 1$

Does this look correct?

**EDIT: Now if I do the same question in another fashion...**

$\displaystyle \frac{1}{1-x} = \frac{1}{1 + (-x)} = \frac{1}{3 + (-x -2)} = \frac{1}{3} \left( \frac{1}{1 + (\frac{-x-2}{3})} \right) = \frac{1}{3} \sum_{n=0}^{\infty} (\frac{-x-2}{3})^{n} = \frac{1}{3} \sum_{n=0}^{\infty} (-1)^{n} \frac{(x+2)^{n}}{3^{n}} = \sum_{n=0}^{\infty} (-1)^{n} \frac{(x+2)^{n}}{3^{n+1}} $ provided that $\displaystyle | \frac{(-x - 2)}{3} | < 1$ or, $\displaystyle -5 < x < 1$

They aren't equal, in this one I have a $\displaystyle (-1)^{n}$ kicking around in my sum.

What am I doing wrong here?