I understand how to find the sum of an infinite geometric series with an n=1
$\sum_{n=1}^\infty a(r)^{n-1} = \frac{a}{1-r}$
However, for sums with N not beginning at 1 (for example, 5 or something else) I was wondering if there was an elegant way to find this sum. I have attempted to subtract the first few terms to no avail. Of course, this assumes that my |r| < 1.
Best Answer
The general formula for the sum of the series is
$$ \sum_{n=N}^\infty r^n = \frac{r^N}{1-r} $$
which can be derived from the one you wrote and the fact that
$$ \sum_{n=0}^N r^n = \frac{1-r^{N+1}}{1-r} $$ which can be proved by induction. Notice that this last formula holds regardless of $r$ (in particular, it holds also for $|r|> 1$).