Finding the sum of series help: $\displaystyle\sum\limits_{n=1}^{\infty} (-9)^nx^n. $ I have a question that I am kind of stuck on. I am good with series stuff, but this one kind of threw me off
$$\sum\limits_{n=1}^{\infty} (-9)^nx^n. $$
This is the question and I was supposed to find out the values of where $x$ converges, and I found that out to be from $-\frac{1}{9} < x < \frac{1}{9}$
How can I find the sum of the series for those values of $x$?

Given the closed form of the geometric series:
$$
\sum_{n=1}^\infty a^n = \frac{a}{1-a}
$$
Plug $a = -9x$ to get:
$$
\sum_{n=1}^\infty (-9)^nx^n = \frac{-9x}{1+9x}
$$
