How to compute $\sum_{n=0}^{\infty}\left(\frac{4}{(-3)^n} - \frac{3}{3^n}\right)$? 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).

$\begin{array}{l}
 \left( {\frac{4}{{\left( { - 3} \right)^n }} - \frac{3}{{3^n }}} \right) = \frac{{\left( { - 1} \right)^n }}{{3^n }} + \frac{{3\left( { - 1} \right)^n }}{{3^n }} - \frac{3}{{3^n }} \\ 
  = \left\{ \begin{array}{l}
 \frac{1}{{3^{2p} }}\quad ;n = 2p \\ 
 \frac{{ - 7}}{{3^{2p + 1} }}\quad ;n = 2p + 1 \\ 
 \end{array} \right. \\ 
  \Rightarrow \sum\limits_{n = 0}^{ + \infty } {\left( {\frac{4}{{\left( { - 3} \right)^n }} - \frac{3}{{3^n }}} \right)}  = \left\{ \begin{array}{l}
 \sum\limits_{p = 0}^{ + \infty } {3^{ - 2p} } \quad ;n = 2p \\ 
  - 7\sum\limits_{p = 0}^{ + \infty } {3^{ - 2p + 1} } \quad ;n = 2p + 1 \\ 
 \end{array} \right. \\ 
 \end{array}$
