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About this Lesson
 Type: Video Tutorial
 Length: 13:16
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 143 MB
 Posted: 06/26/2009
This lesson is part of the following series:
Calculus (279 lessons, $198.00)
Calculus: Sequences and Series (45 lessons, $69.30)
Calculus: The Ratio and Root Tests (3 lessons, $5.94)
Taught by Professor Edward Burger, this lesson comes from a comprehensive Calculus course. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/calculus. The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hopital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.
Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.
He has also taught at UTAustin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".
Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, "Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas" and of the textbook "The Heart of Mathematics: An Invitation to Effective Thinking". He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The "Journal of Number Theory" and "American Mathematical Monthly". His areas of specialty include number theory, Diophantine approximation, padic analysis, the geometry of numbers, and the theory of continued fractions.
Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.
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Sequences and Series
The Ratio and Root Tests
The Ratio Test Page [1 of 3]
So the question now is, how can you figure out if an infinite series converges absolutely? Are there tests that we can actually apply to see if an infinite series converges in this very, very strong way? In a way where not only does the infinite series itself converges, but when everybody pop absolute values around the terms individually, that new thing still converges. Okay, so what kind of test can we apply?
Well, let's just think about this for a second and sort of think about this together. What does it mean for an infinite series to converge? What has to happen? First of all, the individual terms has to shrink down to zero. If the terms don't shrink to zero, there's no hope of this thing converging. But, the next thing I've got to check, is to make sure that in some sense, these terms not only go to zero, but shrink to zero really, really fast, at a very, very fast rate. So, that I can actually do and perform this infinite sum, whatever that means, which, of course, we know what it means. It's the limit of a sequence of partial sum.
I have to make sure that these not only are getting small, but are getting small fast enough. Any way to measure how fast these terms are shrinking compared to the previous terms, for example, is a good measure to see how fast these terms are going to zero. What I'm saying is, suppose that I look at this term and then I look at the very, very next term. And, I see that the next term is a lot, lot, lot smaller than this previous term. Well, then I know that the terms are shrinking faster and faster and faster.
So, in fact, we can take this idea and it's principle, and apply it and what we produce is called the "ratio test". This is a way of seeing if an infinite series converges, by comparing one element in the infinite series to the previous element in the infinite series. And look at that ratio. And, if that ratio is really, really small, then it means that this infinite series, the terms are shrinking so fast that this thing will converge. So, that's sort of the strategy. Now let's see this thing in action.
So, in action, we're looking at this thing, which we call the "ratio test". And, so, how does it work? What we do is, we take a look at this infinite series. And what I'll do is, I'll now compute this quotient, this comparison between adjacent terms in the sum. So, what I consider is the following. Given this, I consider the limit as n goes to infinity of the adjacent terms. So, I take the n plus first term, that's just sort of not the n^th term, but the one that comes after it, and I'll compare it by dividing it by the previous term. So, I'm taking the quotient of two consecutive terms in my infinite series. And, I want to take the limit and I'm going to put absolute values around it, because I do not want to be swayed by negative signs and so forth.
So, I put absolute values around it, take the quotient and I see what that equals. I'll call that "rho", the Greek letter r, it looks like a "p", but it's really a r. Okay, so I look at this thing, r by the way is a Greek letter which is sort of like the "r" and they use that, of course, because it's called the "ratio test". So r, ratio, ratio, r, it all makes sense. Now, you take that r and now what do you do with it? Well, you look at the number. If that number is small, what does it mean? If this number is small, the only way that this limit of this quotient could be small is if, in fact, the top is shrinking faster than the bottom.
And, so, if the top is shrinking faster than the bottom, then that's just enough decrease to make this thing converge. So, if r is small, then the infinite series converges. So, let's write that down now very specifically. So, here is the statement that's the "ratio test". If r ¼ 1, any number ¼ 1, then what can we say about this infinite series? Then, in fact, not only does the infinite series converge, but it converges absolutely. This thing actually converges absolutely. So, not only does it converge, but if you put absolute values around it, it still is going to converge. So, it converges in a very, very strong way as long as this ratio limit is less than 1, anything less than 1. It could be zero, anything as long as it's less than 1.
Okay, fine, so that's great. So then converges, then summation one over n, converges absolutely. Great. All right, now what about if the r is big? Well, if the r is big, that means that in some sense this n plus first term, is not shrinking fast enough. And, if it's not shrinking fast enough, this thing should diverge. So, intuitively, it seems reasonable to say that if r ¾ 1, then what do we say? Well, then we say the infinite series diverges, that's the end of it. So, if the ratio is big, then, in fact, the terms aren't shrinking fast enough, and so we have divergence. So then the infinite series diverges. It converges absolutely if it's less than 1, bigger than 1 it diverges.
Now, there's one particular value for r that we haven't actually mentioned yet, so let's mention that. There's one number we haven't thought about. What happens when it actually equals 1? Well, then what can we say about the infinite series? Well, then, we can say about the infinite series is actually not much. Because in that case, what happens is the test is inconclusive. That's it, game over, we've learned nothing. And, so, what that means is, this thing might converge, it might diverge, it might converge absolutely, it might converge conditionally, we just don't know. This test failed, inconclusive.
So, it's really important to understand when a particular test just is inconclusive. Many people have already had the classic mistake as to get that r = 1, and then make some conclusion from that. Oh, therefore, the thing converges, or therefore, the thing diverges. You can't do that. If r = 1, all you can say is, "I shouldn't have used the ratio test." That's all you can say. You have no idea about what's going on with respect to the original infinite series at hand. So, in this case, then game over, test inconclusive. So, it needs more work.
Okay, so that's the idea of the ratio test, and it's actually something that I want to have it make sense to you. You take a look at the ratio and now the order is very important, by the way. I take a big term and divide it by the previous term. And I take the absolute value and look at the limit. If that limit is small, namely less than 1, then I know that the infinite series, the associated infinite series converges, in fact, converges absolutely. If, in fact, that ratio is large, that means the terms are not shrinking fast enough, so in that case it diverges. And, if the ratio is 1, then we know nothing, we know absolutely nothing.
So, let's take a look at how we can use the ratio test here. It's a great test by the way, I happen to really like the ratio test. So, let's take a look at the infinite series, and going from one to infinite of one over 5 to the n. Now, what is the answer? Does this converge or diverge? Well, actually, I happen to know the answer, and I hope that you do to, because you may recognize this is a geometric series. This is a geometric series where the ratio, where the individual "R" here is going to be 1/5, 1/5 ¼ 1. So, in fact, this not only converges, but converges absolutely, since there are no negative signs, putting absolute values didn't change anything. This converges absolutely and, in fact, this is one of the few infinite series where we can actually say what the sum is. Because we know there is a formula for the actual sum of a geometric series.
Anyway, but putting all that aside, let's actually apply the ratio test to verify what we already know. So, how does it work? I take the absolute value of the n plus first term divided by the n^th term. Now, what does that equal in this case? Well, in this case, I've got to look at a^n+1. So, that means I look at n plus first term here, so this n becomes n+1. So, wherever I see an n, I put an n+1. So I see 1 over 5^n+1, that's this term. Now I have to divide that by this term, which is one over 5^n. Now I put absolute values there but, of course, since everybody is already positive, they can just evaporate.
Now, this is a compound fraction, so let's invert and multiply, take this, watch what happens to the 5^n, comes up on top. So, I see the limit as n goes to infinite of 5^n divided by 5^n+1. And there's a lot of cancellation here. In fact, what am I left with? I can cancel the 5^n away, but I'm left with one extra 5 downstairs, this is just 1/5. And what's the limit of 1/5? It's just 1/5. So r in this case, equals 1/5. So that ratio limit is equal to 1/5. Notice 1/5 ¼ 1, so this thing converges and converges absolutely by the ratio test. So this converges absolutely. So there's a neat example confirming what we had already known.
Okay, let's try another one. How about summation n going from one to infinite of 1.1^n. Another example of a geometric series, but notice that here the geometric series, the "R" in the geometric series is actually a number greater than 1. And we've already seen that if you have a number greater than 1, and what's happening is, this thing must diverge. So we know the answer in advance, it's another one that we already know, this is diverging. Let's see if the ratio test will confirm that for us. So, to apply the ratio test, I take the absolute value of the n plus first term, divided by the n^th term. Why does the n plus first term come first? I want to see how fast they're shrinking, so I take one term and then divide it by the previous term. So the order is always of this flavor. And what happens if you insert all the data? Well, I would put in 1.1^n+1 divided by the n^th term, which is 1.1^n. And what happens here? Well, there was a lot of cancellation and, in fact, all I'm left with is 1.1^1, and that limit is 1.1. So, in this case, r is just 1.1, which notice is a little bit bigger than 1, not much bigger, but all I'm asking is, is it bigger than 1? Since it is, then what I know is, this must diverge. So, by the ratio test I can confirm what we already knew, this diverges, no hope of conversion at all.
Okay, how about this one? Summation one over n^2, n going from 1 to infinite. So, let's try the ratio test, limit as n goes to infinite of a^n+1 over a^n, an absolute value. So what does that equal? Well, the n plus first term would be 1 over n+1 all squared. Right? Because this whole thing is squared, so the next term would be 1 over n+1^2. And I divide that by the n^th term, and I put absolute values here, but notice again, everything is positive here, so I just drop them immediately. If there are negative terms here, you got to clean that up. When you invert and multiply, I'm left with this. And what's that limit? Well, one application or two applications of L'Hospital's rule, will immediately allow you to verify this limit as 1. Or, you could just notice that the growth rate on top is n^2, the growth rate on the bottom is n^2, the coefficients are both 1, the limit is 1.
One is r, and that equals 1, which means that the ratio test here is actually inconclusive. So, we don't know if it converges or diverges by the ratio test. However, we do know the answer, because if you notice, this is just a pseries where p=2. So, in fact, we know this does converge. So, here's an example of an infinite series that converges absolutely  no negative signs  and yet the ratio test is just as inconclusive. So, it needs more work in this case, of course, we already know the answer.
And, let's do one final one, big finish. Well, not a big finish, but certainly a finish. I guess all finishes are big if you promise to finish. Okay, how about one over n? Well, let's take the limit as n goes to infinite of the absolute value of a^n+1 over a^n. And what does that equal? That equals the limit and then goes to infinite. The n plus first term is 1 over n+1, divided by the n^th term, which 1 over n, no absolute values, because everything is positive. And, when I invert and multiply, I get the n on top, n+1 on the bottom. One application of L'Hospital's rule, I see the answer is 1. So r=1, and so 1= r. Which means again, the ratio test is inconclusive, and so this needs more work.
But, we know this is the harmonic series which we already proved using the integral test that, in fact, this diverges. So, here's an example where the test is inconclusive and the thing diverges. The previous example was an example where the test actually was inconclusive and the thing converged. So, you can see both possibilities are possible. That's why they're called possibility.
So, be very careful. If you get a r that equals 1, make sure you do something else and conclude nothing, the ratio test just wasn't a good test. However, the ratio test allows us to actually find out whether things converge or diverge in a very simple way, if we can compute that particular limit, and if it doesn't equal 1, than we're in hog heaven. See you at the next lecture.
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