# Trying to find interval of convergence for a geometric series

1. Mar 6, 2012

### skyturnred

1. The problem statement, all variables and given/known data

here is the series:

$\sum^{\infty}_{n=0}$x(-15(x$^{2}$))$^{n}$

2. Relevant equations

3. The attempt at a solution

I know that

-1<-15x$^{2}$<1 for convergence (because of geometric series properties)

but I run into a problem here:

-1/15<x$^{2}$<1/15

You can't take the square root of a negative number.. so is the interval of convergence just

[0,(1/15)$^{1/2}$)?

Thanks

Last edited: Mar 6, 2012
2. Mar 6, 2012

### LCKurtz

No. The left side of that inequality is free so you just have $x^2<1/15$. When you take the square root of both sides remember $\sqrt{x^2}=|x|$ so you get $|x|<1/\sqrt{15}$ which is $-1/\sqrt{15}<x<1/\sqrt{15}$.

3. Mar 8, 2012

OK Thanks!