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I am trying to find the interval of convergence, and represent a function as a power series. I found the interval of convergence to be (-2,2). I have tried both differentiating, and then taking the integral of its results, and going with $$r=\frac{-x}{2}$$ however, I am unable to get the correct answer for the power series representation. The function is: $$f(x)=\frac{(x - 1)}{(x + 2)}$$ The correct answer is given as: $$-\frac{1}{2}-\sum_{n=1}^ \infty \frac{(-1)^n 3x^n}{2^{n+1}}$$

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Notice that

$$f(x)=\frac{x-1}{x+2}=\frac{x+2-3}{x+2}=1-\frac3{2-(-x)}=1-\frac{3/2}{1-(-\frac x2)}=1-\frac32\sum_{n=0}^\infty(-1)^n\left(\frac x2\right)^n$$

Due to the geometric series, where it converges for $\left|\frac x2\right|<1\implies|x|<2$ by the term test.

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    $\begingroup$ That should do it. +1 $\endgroup$
    – Mark Viola
    Nov 27 '16 at 1:26
  • $\begingroup$ I added what the answer is supposed to be. Yours looks the similar, except it has -1/2 and not 1. Also, how did you notice that the function went into the form 1- (3/2)/1-(-x/2)? $\endgroup$
    – Fyree
    Nov 27 '16 at 2:00
  • $\begingroup$ @Fyree Your answer key moved the 3/2 inside the sum, then it took the $x^0$ out. $\endgroup$ Nov 27 '16 at 2:08
  • $\begingroup$ @Fyree And I've updated the answer. Notice that the second to last step just divides the fraction by $2$. $\endgroup$ Nov 27 '16 at 2:10
  • $\begingroup$ I finally get it now. Thank you for the help! $\endgroup$
    – Fyree
    Nov 27 '16 at 2:34

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