For what values of r would the series converge?

Question

Find the values of r for which $\sum_{k=0}^\infty 3^kr^{2k}$

I understand that the common ratio is 3$r^2$ so I know that it converges if and only if |3$r^2$| is less than 1.

So:

-1 < 3$r^2$ < 1

$-\frac{1}{3}$< $r^2$< $\frac{1}{3}$

And this is where I get stuck.

Any suggestions on how to continue?

Answer 1

Hint. Use that $$ 0\le r^2<\frac13 \iff \left(r-\frac1{\sqrt{3}} \right)\left(r+\frac1{\sqrt{3}} \right)<0\iff -\frac1{\sqrt{3}}

Answer 2

You're almost there. Just recall that if $x^2

Here, we have $x^2=3r^2$ and $x_0^2=1$. Therefore,

$$|\sqrt{3}\,x|<\sqrt{1}\implies |x|<\sqrt{\frac13}\implies -\sqrt{\frac13}