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# Worked example: limit comparison test

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.9 (EK)

## Video transcript

so we're given a series here and they say what series should we use in the limit comparison test let me underline that the limit comparison test in order to determine whether s converges so let's just remind ourselves about the limit comparison test if we say if we say that we have two series and I'll just use this notation a sub N and then another series B sub N and we know that a sub N and B sub n are greater than or equal to 0 for all n for all n if we know this then if then if the limit as n approaches infinity of a sub n over B sub n is equal to some positive constant so 0 is less than that constant is less than infinity then either both converge then both converge both converge both converge or both diverge both diverge and it really makes a lot of sense because it's saying look as we get into our really large values of n as we go really far out there in terms of the terms if our behavior starts to look the same then it makes sense that both these series would converge or diverge and we have an introductory visio we have an introductory video on this in another video so let's think about who what if we say that this is our a sub n if we say that this is a sub n right over here what is the series that we can really compare to that seems to have the same behavior as n gets really large well this one gets to seems to gets unbounded this one doesn't look that similar has a 3 to the N minus 1 in the denominator but the numerator doesn't behave the same this one over here is interesting because we could write this this is the same thing as the sum N equals 1 to infinity we could write this as 2 to the N over 3 to the N and these are very similar the only difference between this and this is that in the denominator here or in the denominator up here we have a minus 1 and down here we don't have that minus one and so it makes sense given that that's just a constant that as n gets very large that these might behave the same so let's try it out let's find the limit and we also know that the a sub NS and the V sub ends if we say that this right over here is V sub n we say that's B sub n that this is going to be positive or this is gonna be greater than or equal to 0 for N equals 1 to 3 so for any values this is going to be greater than or equal to 0 and the same thing right over here it's going to be greater than or equal to 0 for all the ends that we care about so we meet these these first constraints and so let's find the limit as n approaches infinity of a sub n which is I'll write in that red color which is 2 to the nth power over 3 to the n minus 1 over B sub n over 2 to the N over 3 to the N so let me actually do do a little algebraic manipulation right over here this is going to be the same thing as 2 to the N over 3 to the n minus 1 times times 3 to the N over 2 to the N divided the numerator and denominators by 2 to the N those cancel out and so this will give us this will give us 3 to the N over 3 to the N minus 1 agree we can divide the numerator and the denominator by 3 to the N and that'll give us 1 over 1 minus 1 over 3 to the N so we could say this is the same thing as the limit as n approaches infinity of 1 over 1 minus 1 over 3 to the N well what's this going to be equal to well as n approaches infinity this thing 1 over 3 to the N that's going to go to 0 so this is this whole thing is just going to approach 1 and one is clearly between 0 and infinity so the destinies of these two series are tie they either both converge or they both diverge and so this is a good one to use the limit comparison test with and so let's think about it do they either both converge or do they both diverge well this is a geometric series our common ratio here is less than 1 so this is going to converge this is going to converge and because this one converges by the limit comparison test our original series s converges converges and we are done
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