Convergence of a geometric sequence

I am not sure if there is such thing as testing for convergence a geometric sequence, but I have a problem in my book that asks to test for the convergence or divergence of a sequence:

$\lim\limits_{n \to \infty} \frac{2^n}{3^{n-1}}$

Can I use the same method used to test a geometric series for this sequence?

So rearranging:

$\lim\limits_{n \to \infty} 2(\frac{2}{3})^{n-1}$

So in this case a=2 and r=$\frac{2}{3}$ so my sequence would converge as my r is within -1 and 1?

Is this thinking wrong? If so what method should I be applying to solve this?

Thank you!

1 Answer

For any geometric series, if $|r|<1$, then your series will converge. Your reasoning is perfectly sound. If a series converges, then the limit of its corresponding sequence is zero.