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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!

melm
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1 Answers1

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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.

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  • However the question was asking about the sequence. It didn't state that it was a series. So I can apply the test for a geometric series to a geometric sequence? (Is there such thing as a geometric sequence?) – melm Dec 03 '16 at 04:06
  • Yes you can. If the series converges, the sequence must also converge. – Anthony P Dec 03 '16 at 04:13
  • Thank you both!! I understand now, very helpful. – melm Dec 03 '16 at 04:15