Finding the sum of series help: $\displaystyle\sum\limits_{n=1}^{\infty} (-9)^nx^n. $

Question

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 $x$?

Answer 1

Given the closed form of the geometric series: $$ \sum_{n=1}^\infty a^n = \frac{a}{1-a} $$

Plug $a = -9x$ to get: $$ \sum_{n=1}^\infty (-9)^nx^n = \frac{-9x}{1+9x} $$

Answer 2

To compute the interval of convergence for a power series $\sum a^n x^n$, compute $$\lim_{n\to \infty} \sqrt[\large n]{|a_n|} = \lim_{n\to\infty} \sqrt[\large n]{9^n}$$

In this case, that gives us $9$, so the radius of convergence is $\dfrac {1}9$ which will converge when $|x| \lt \dfrac 19$

Answer 3

It can be written as $(-9x)^n$ so the ratio here is $-9x$. Since it starts at $1$ so the first term is $-9x$ and that is also the ratio, so $\frac{-9x}{1+9x}$ because you use the formula (1st term)/(1-ratio) so $x$ will still have to be in your answer.