Finding the Laurent Series for 1/(x+3) around x=2

[MrGandalf]

Homework Statement

I know the sum of the Laurent series (around x=2) is equal to
$$\frac{1}{x+3}$$
But I can't find what the series is from this information alone.

Homework Equations

In the textbook, you have (for -1 < x < 1):
$$\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$$
and for |x|>1 I know (but have no idea how to deduce) that
$$\frac{1}{1+x} = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{x^n}$$

I just don't know how I can use this information to find the sum for 1/(x+3).

The Attempt at a Solution

I am sorry, but I don't want to further destroy my confidence by reliving my pathetic attempts to finding the solution. :D

 

[dirk_mec1]

Yes you can, you just use the geometric series

$$\frac{1}{x+3}=\frac{1}{3-(-x)} =\frac{1}{3(1- \frac{-x}{3})}$$

Now use u = -x/3 and employ the geometric series for |x|<1.

 

[MrGandalf]

Thank you! A clever little move there.

I'll be sure to include you in my 'Thank You' speech when I accept my Fields medal. ;)

 

[Dick]

If you want the Laurant series around x=2, you want a series of powers of (x-2). You might want to rearrange the form a bit before you do the geometric series trick.