I have numerically verified that the following sum comes to $\displaystyle \frac 94$, the same thing the back of the book says. So what'd I do wrong?

Find the sum of the convergent series

$\displaystyle 3-1+\frac 13 - \frac 19 + \mathellipsis$.

I verified that the above sum can be represented as

$\displaystyle \sum_{n \ge 1} \frac{(-1)^{n+1}9}{3^n}$,

which can be rewritten as

$\displaystyle \sum_{n \ge 1} \frac{(-1)^{n+1}9}{3^n} = -9\sum_{n \ge 1} \frac{(-1)^{n}}{3^n} = -9\sum_{n \ge 1} \left(\frac{-1}{3}\right)^n$.

I believe I can apply the geometric series to the summation, giving me

$\displaystyle -9\sum_{n \ge 1} \left(\frac{-1}{3}\right)^n = -9 \cdot \frac{1}{1+\frac 13} = -9 \cdot \frac 34 = -\frac{27}{4} \ne \frac 94$.

Ahhh... but now I see that the geometric series equation starts the summation at $\displaystyle n=0$ (do you find that typing up your answers sometimes helps you realize what's wrong? Why is that?).

So, I need to subtract $\displaystyle -9 \cdot \frac{-1}{3}^0 = -9$ from my answer, which (unsurprisingly) gives $\displaystyle \frac 94$.