Trying to find interval of convergence for a geometric series

[skyturnred]

Homework Statement

here is the series:

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

Homework Equations

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

 

[LCKurtz]

f 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

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 $-1/\sqrt{15}<x<1/\sqrt{15}$.

 

[skyturnred]

OK Thanks!