1. ## Sigma Notation

Consider the series the sum of x^-n for values of n from 1 to infinity.
Looks kinda like that:

infinity
sigma x^-n.
n=1

a) What is the interval of convergence?
b) In terms of x, find the sum of the series.

I have no idea how to find the interval of convergence
I tried making charts where x =1 and I'd go
n = 1, x^-n = 1
n = 2, x^-n = 1/1^2 = 1
n = 3, x^-n = 1/1^3 = 1
....
...
and then x = 2
n = 1, x^-n = 1/2
n = 2, x^-n = 1/2^2 = 1/4
n = 3, x^-n = 1/2^3 = 1/8
...
...
I'm assuming the interval of convergence is when my outputs start becoming positive and negative. (ex: -1, 2, -3, 4...)

Then for B, I tried using the sequence formula of adding the sum of an infinite geometric sequence.
I did: Sn = T1 / 1-r
= X^-1 / 1-(-n)
= 1/x / 1+n
= 1/x * 1+n/1 = 1+n / x

Help would be greatly appreciated, thank you!

~Wingless

2. $\displaystyle \sum_{n=1}^{\infty}\frac{1}{x^n}$

if -1<x<1 how the series will be ??

$\displaystyle \left(\frac{1}{\frac{1}{2}}\right)^{n}=??$

if you know how it will be when -1<x<1 and when x in [1,infinity) or (-infinity,-1]

use ratio or root test then you will get the convergence interval