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I'm currently trying to compute the following series (found on page 65 of this textbook):

$$\sum_{n=0}^{\infty}\left(\frac{4}{(-3)^n} - \frac{3}{3^n}\right)$$

I've tried to somehow transform it into a geometric series (which I'm fairly sure is the strategy for this series), but I've been unable to. Any help in solving this would be appreciated (though I'd prefer a hint over a solid answer).

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Hint:

$$\frac{4}{(-3)^n} - \frac 3 {3^n} = 4 \left(- \frac 1 3\right)^n - 3 \left(\frac 1 3\right)^n$$

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  • $\begingroup$ Hold up - can I distribute the summation to both of those terms? If so, this question is much easier than I thought. $\endgroup$ – Cisplatin Mar 14 '14 at 0:30
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    $\begingroup$ @Sim Yes: Both series are absolutely convergent, so you can sum separately. $\endgroup$ – user61527 Mar 14 '14 at 0:31
  • $\begingroup$ If the downvoter would suggest a correction or an improvement, I'd appreciate it. $\endgroup$ – user61527 Mar 14 '14 at 1:04
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$\begin{array}{l} \left( {\frac{4}{{\left( { - 3} \right)^n }} - \frac{3}{{3^n }}} \right) = \frac{{\left( { - 1} \right)^n }}{{3^n }} + \frac{{3\left( { - 1} \right)^n }}{{3^n }} - \frac{3}{{3^n }} \\ = \left\{ \begin{array}{l} \frac{1}{{3^{2p} }}\quad ;n = 2p \\ \frac{{ - 7}}{{3^{2p + 1} }}\quad ;n = 2p + 1 \\ \end{array} \right. \\ \Rightarrow \sum\limits_{n = 0}^{ + \infty } {\left( {\frac{4}{{\left( { - 3} \right)^n }} - \frac{3}{{3^n }}} \right)} = \left\{ \begin{array}{l} \sum\limits_{p = 0}^{ + \infty } {3^{ - 2p} } \quad ;n = 2p \\ - 7\sum\limits_{p = 0}^{ + \infty } {3^{ - 2p + 1} } \quad ;n = 2p + 1 \\ \end{array} \right. \\ \end{array}$

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