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Obtain power series expansion of $\frac{1}{(1+x)(1-2x)}$ and give the general term and radius of convergence.

user2250537
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  • You've given the general term, but not a series. Your term is not correct - how did you get it? –  Mar 31 '16 at 21:19
  • We only have convergence for $|x|\lt 1/2$. First use partial fractions on $\frac{1}{(1-x)(1+2x)}$. – André Nicolas Mar 31 '16 at 21:22
  • @user1150537 How can the series possibly converge to $f(x)=\frac1{(x+1)(1-2x)}$ for $-1<x<1$? What value would it take in $x=\frac12$? –  Mar 31 '16 at 21:49

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Splitting into partial fractions, you get $$\frac{1}{(1+x)(1-2x)}=\frac{1}{3}\left(\frac{1}{1+x}+\frac{2}{1-2x}\right)$$ If you're familiar enough with the series expansion for $\dfrac{1}{1-x}$, it should be easy to see that the series expansion in this case would simply be $$\frac{1}{3}\left(\sum_{n\ge0}(-x)^n+2\sum_{n\ge0}(2x)^n\right)$$ which can be consolidated into one series: $$\frac{1}{3}\sum_{n\ge0}\bigg((-1)^n+2^{n+1}\bigg)x^n$$ As for the radius of convergence, recall that the geometric series $\sum\limits_n x^n$ converges for $|x|<1$. You should be able to establish for yourself that in general, $\sum\limits_n (ax)^n$ converges for $|ax|<1$, or $|x|<\dfrac{1}{|a|}$. Then the R.O.C. must be $\dfrac{1}{|a|}$. This means the R.O.C. for $\sum\limits_n (-x)^n$ is $1$, while the R.O.C. for $\sum\limits_n (2x)^n$ is $\dfrac{1}{2}$. Note that both series converge only for $|x|<\dfrac{1}{2}$.

Alternatively, you can use the more compact form of the series expansion and apply the ratio test.

user170231
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