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# Ratio test

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.11 (EK)

## Video transcript

we already have a lot of experience with the geometric series for example if I have the infinite geometric series starting at N equals K to infinity of R to the N which would be R to the K plus R to the K plus 1 plus R to the K plus 2 and keep going on and on and on forever there's a few things we've already thought about here we know what the common ratio is so the common ratio the common ratio which is the ratio between consecutive terms is going to be R to the K plus 1 actually let me just write this as R to the n plus 1 I don't want to fixate on this K right over here so R to the N plus 1 over R to the N well this is just going to be equal to R anything to the n plus 1 over that same thing at the end that's just going to be equal to R R to the first power and you see that here when you go from one term to another you are just multiplying you're just multiplying by R and this is all this is all review if it's not review I encourage you to watch the videos on geometric series and what's interesting about this is we've proven to ourselves in the videos on geometric series that if if if the common ratio if the absolute value of the common ratio is less than 1 then then the series the series converges converges and if the absolute value of R is greater than or equal to 1 then the series diverges and that made sense we've proven it as well but it also makes logical sense that look if if the absolute value of R is less than 1 then each term here is going to be is going to go down by that common or it's going to be multiplied by that common ratio and it's going to decrease on and on and on and on and on and so it makes sense that even though this is an infinite sum it will converge to a finite value now with that out of the way for review let's tackle let's tackle something a little bit more interesting let's say that we want to look at a we want to figure out whether a series like this so starting with N equals 5 to infinity of let's say n to the 10th power the numerator is growing quickly n to the 10th power over n factorial n factorial we know also grows very very quickly probably it actually grows much faster than then even then even a high degree polynomial like this or a high degree term like this but how do we prove that it converges we can definitely use the divergence test to show that this does not diverge but how do we prove that this actually converges and maybe we can use a little bit of our intuition here well let's see if we can come up with a common ratio so let's see if there's a common ratio here so let's take the n plus 1 term which is going to be n plus 1 to the 10th power over n plus 1 factorial and divide that by the end term so let's divide that by n to the 10th over n factorial well dividing by a fracture divided by anything is equivalent to multiplying by its reciprocal so let's just multiply by the reciprocal so times n factorial over N to the 10th remember all I'm trying to do is exactly what we did right over here see if there's some type of a common ratio well if we do a little algebra here n plus 1 factorial n factorial algebra is always fun this is the same thing as n plus 1 times n factorial times n factorial you're not used to seeing order of operations do them with factorial where the spectral only applies to this n right over here and why is that useful because this n factorial cancels with that n factorial and we're left with n plus 1 to the 10th power over n plus 1 times n to the 10th times n to the 10th so I know what you're thinking hey wait look this isn't a fixed common ratio here the ratio between consecutive terms here when I took the n plus 1 term divided by the nth term it's changing depending on what my n is this ratio I guess you could say is a function is a function of n so this doesn't seem too useful but what if I were to say okay well look in any of these series we really care about the behavior as our ends get really really really real our jazz we the limit as our ends go to infinity so what if we were to look at the behavior here and if this is approaching some actual values and approaches infinity well it would make conceptual sense that we could kind of think of that as our what the common the limit of our common ratio so let's do that so let's take the limit as n approaches infinity of this thing so remember what we're doing here this isn't just kind of some voodoo conceptually so let me copy and paste this conceptually we're just trying to think about well what is the common ratio approach what is the ratio between consecutive terms the n plus 1 term in the nth term as n gets really really really really really large and what you see here is we have this up here we don't have to multiply it out this is going to be n to the 10th plus a bunch of other stuff it's going to be a 10th degree polynomial and this down here this is going to be actually this one we can figure it out this is going to be n to the xi plus n to the 10th power so the limit as n approaches infinity well this denominator is going to grow faster than that you could divide the numerator and denominator by n to the 10th and all of and and everything well you could divide actually the numerator denominator by n to the 11th and everything up here is going to go to 0 and this is going to be 1 and so the limit right over here is the limit this limit is going to be equal to this limit is going to be equal to 0 as n approaches infinity one way the ratio between the ratio between consecutive terms approaches zero so this seems if you if we use the the the logic from our common ratio up here when we looked at geometric series this is clearly not a geometric series but it would say well look as n gets really really really large the ratio between consecutive terms gets smaller and smaller and smaller hey maybe we can make the same conclusion maybe we can make the same conclusion that therefore the series the series actually converges the series actually converges so or n factorial converges so can we make this statement and the answer is yes we can and what allows us to do it is the ratio test is the ratio test let me write that down the ratio test and all the ratio test tells us is if we have an infinite series if we have an infinite series and I can say from N equals K to infinity and if we say if we take the limit as n approaches infinity as n approaches infinity of the absolute value of a to the n plus 1 over a to the N and over here I didn't take the absolute value of that limit but we absolutely could have taken the absolute value of that limit and these were all of these terms right over here were positive so the absolute value would be the same thing as itself if this approaches some limit if this approaches some limit so once again this limit is what is going to be our R ratio between consecutive terms as n gets larger and larger the ratio tells us if if L is less than 1 which was the situation here L is clearly less than 1 then the series converges series converges some definitions you say they'll it's absolutely convergent will convert absolutely the absolute value of the series converges but then that also means that the series itself converges if L is greater than 1 series diverges series diverges and if L is equal to 1 its inconclusive we don't know we don't know we would have to do some other test to prove whether it converges or diverges so that's the essence of the ratio test find the absolute value of the ratio between consecutive terms take the limit as n approaches infinity if that approaches a actual limit and that limit is less than 1 then the series converges and it's really based on the same fundamental idea that we saw with the common ratio of geometric series