$\sum_{n=1}^\infty \frac{1}{|x|^n}$ does it converge for |X| >1?

Question

i have been dealing with this series, i tried the root test on it to check if it converge or diverge, and i received that in order for the series to converge |X| should be lower than one, i.e $|X|<1$ . It does not seems right to me since lets say i will plug in 100 than it suppose to diverge, but this series for $x=100$ converge.

i can show it : let $B_n = \frac{1}{n^2}, \sum_{n=0}^\infty b_n $ known as a converge. $\lim_{n\to\infty} \frac{a_n}{b_n}$ $\lim_{n\to \infty}\frac{n^2}{100^n}=0 $ so by that $\sum_{n=1}^\infty \frac{1}{100^n}$ converge as well. where is my mistake? thanks.

Answer 1

If $a_n=\frac 1 {|x|^{n}}$ then $a_n^{1/n}=\frac 1 {|x|} <1$. Root test tells you that the series converges. You can also apply ration test.

Answer 2

By the geometric summation formula,

$$\sum_{n=1}^m\frac{1}{|x|^n}=\frac{1-\dfrac{1}{|x|^{m+1}}}{1-\dfrac{1}{|x|}}-1$$

and this converges for $\dfrac1{|x|}<1$. The limit is $\dfrac1{|x|-1}$.

Answer 3

Let $|x|>1$.

$|x|=1+y$, $y >0$.

$|x|^n =(1+y)^n \gt $

$1+ ny +[n(n-1)/2]y^2+.....+ \gt [(n-1)^2/2]y^2$.

Hence

$\dfrac{1}{|x|^n} \lt (\dfrac{2}{y^2}) \dfrac{1}{(n-1)^2}$.

Comparison test.