 Determine the convergence of the given series. The sum, where n ranges from 1 to infinity, of negative 2 to the n over 3 to the n minus 1. Maybe pause this video for a second and consider how would I determine the convergence or divergence of this series, which test would I use? Now for me, I would actually treat this as a geometric series. So consider the following. If I take this sum and times it by 3 over 3, keeping still the sum from 1 to infinity, you have your negative 2 to the n over 3 to the n minus 1. The reason I want to do this, I mean times it by 3 over 3 of course is just the number 1, right? But the advantage here is if I combine this 3 with the 3 minus n's we already have in the denominator, then this series will become 3 times the sum of negative 2 over, well I'll back up for a sec. We're going to get negative 2 to the n over 3 to the n as n goes from 1 to infinity. And therefore rewriting this just one more time, admittedly we could have gotten this done really quick, and equals 1 to infinity we have negative 2 thirds to the n. And so now we can actually see this series as the geometric series that I claimed it was. It's a geometric series. And the convergence of a geometric series comes entirely from its constant ratio right here. We see that r is going to equal negative 2 thirds. And a geometric series is convergent exactly when its ratio is small. That is the absolute value of r is strictly less than 1, which it is in this case. Therefore we should declare that this series is convergent. It's in fact convergent by the geometric series test because it has a small ratio. So just this slight little modification here, and why did I do that in the first place? I noticed looking at this series that I had a power of n for the negative 2 and I had some exponential expression of 3 there. Since I had these exponential expressions going on with my series, it made me think that okay, it's going to be a geometric series and this one is convergent because its ratio is small.