interval of convergence

Question

anonymous · Answer

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

anonymous · Answer

i used partial fractions to get$$\frac{ 1 }{ (x-2) }-\frac{ 1 }{ (x+1)}$$

anonymous · Answer

using the geometric series representation i got this..$$-\frac{ 1 }{ 2 }\sum_{n=0}^{\infty} (\frac{ x }{ 2})^n-\sum_{n=0}^{\infty}(-x)^n$$

anonymous · Answer

my question is why is the interval of convergence of sum 1. because with this series you would normally get 2 intervals |x|<2 and |x|<1, but why is the interval of convergence of sum is only |x|<1?

anonymous · Answer

sorry i mean the interval of convergence is -1<x<1

ganeshie8 · Answer

I think when x is between 1 and 2, the second sum diverges

anonymous · Answer

ohhh thanks! :D

ganeshie8 · Answer

http://www.wolframalpha.com/input/?i=%5Csum+%5Climits_%7Bx%3D3%7D%5E%7B%5Cinfty%7D+%5Cfrac%7B+3+%7D%7B+x%5E2-x-2%7D do u knw why the interval of convergence is -13 ?

anonymous · Answer

i think so is it related to the geometric series? when r>= 1 it diverges and r<1 it converges?

ganeshie8 · Answer

lol idk... ive never worked on series before...