Originally Posted by

**jegues** 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?