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Calculus: The Root Test

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About this Lesson

  • Type: Video Tutorial
  • Length: 13:29
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 145 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 UT-Austin 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, p-adic 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 Root Test Page [1 of 3]
So the name of the game is to figure out whether an infinite series converges or diverges, and one powerful technique and one powerful test that we can use is the ratio test. Take the ratio of two consecutive terms in the infinite series, a term divided by the previous term, take the absolute value, take the limit. If that limit is small, namely less than one, the thing converges absolutely. If it is big, that limit is actually greater than one, it diverges. And if it equals one the test is inconclusive.
There's another test that's very similar to the ratio test, but allows us to easily take a look at an infinite series and determine whether it converges absolutely or diverges similar to the ratio test. In fact, it's almost like a twin. It's a great idea whenever you see an infinite series that has powers within. Now, if you see a power, like 2^n or something like that, the ratio test will often work. But if you see a really complicated thing, which actually has and the exponent ends, then think something else. What you should think is, "Well, how do you undo if you have an n power?" Take the n^th root. Welcome to the n^th root test. So the n^th root test is a way of extracting n's that are powers. By taking n^th roots, they just go away.
Now let me reason with you on this one so you can see the idea, the theme, just as we saw on the ratio theme. If this thing has to go to zero at very fast clip, one way to measure that is to take the n^th root and ask what happens to the n^th root. If that thing actually shrinks to zero really fast, then what that means is that the thing without the n^th root, in fact, must be really shrinking to zero and so this thing converges very quickly. Therefore, we think that the whole infinite series must, in fact, converge. So the idea is if you take the n^th root in some sense trying to actually prevent the thing from going to zero really fast, and it still goes to zero pretty fast, then it converges.
So what exactly is the n^th root test? Well, the n^th root test says the following. Suppose you have this infinite series and we now construct a new limit. So I'm going to still call it limit, n going to infinity, and now what I'm going to do is instead of taking a ratio of terms, I'm just going to take one term, a[n], take the absolute value of it, so strip away negative signs, and then take the n^th root of that, and then take the limit. I'll call that limit rho. Rho for root this time. We call it rho, but now think about rho as a ratio, but as a root of the term.
Well, the test actually looks the exact same, has the exact same feel to it. If, in fact, the rho is small, less than one, then I say that the infinite series converges. And not only does it converge, it converges absolutely. Then infinite series converges absolutely.
And what if the ratio is large? If the n^th root is greater than one, then as we saw with the ratio test, this will diverge. This is not going to zero fast enough. The terms are not going to zero fast enough, so then the infinite series diverges.
Leaving one case left, what if rho equals one? Well, then as always, we have a game over. This test was inconclusive. The thing might converge, the thing might diverge, the thing might converge absolutely, the thing might converge conditionally. We just don't know. This test fails. It can conclude nothing.
So that is the n^th root test. You can use the n^th root test whenever you see things that are appearing with n powers. So let's take a look at some examples.
Let's take a look at the infinite series, . Now again, I'm starting with a familiar one. This happens to be a geometric series where the r is 1/2, and so I know this actually converges, it converges absolutely since there's no negative signs. And in fact, moreover, I also know that this actually converges to a particular target that I can compute, because geometric series we actually know what the sum equals.
So let's verify all that using the n^th root test. Let's just see the n^th root test in action. So I take the limit as n goes to infinity of the n^th root of the absolute value of the n^th term. What does that equal in this case? That equals the limit as n goes to infinity of the n^th root of - well, the absolute value of this is just itself. So I see here is just . Now happens if you take the n^th root if this? Well, the n^th root of 1 is 1. And what's the n^th root of 2^n? Well, the n^th root and the n here, they annihilate themselves, and it's left with 2. And what's that limit? That limit is a half. Notice that that half, therefore, is the rho, is the root limit. That is less than one, and so therefore, we know by the root test that this thing converges absolutely, just as we already know and which was confirmed by the root test. This converges absolutely by the n^th root test. An easy example, but I want us to get in the habit of seeing how this limit looks.
Well, let's try another one. . This is also another geometric series where now the r is actually -3, which you notice in absolute value is bigger than one. So we're already seen this should diverge. Let's verify that right now by using the root test. So if you use the root test we take the n^th root . What does that equal in this case? What's the limit, and then goes to infinity of the n^th root of what? Of the . And what does that equal? Well, the absolute equal of - and that's -3. It's all right at the end. So the (-1)^n and absolute value just becomes a 1, because negative and positive, you throw those away. So what I see here is just 3^n. So this is the limit as n goes to infinity of the n^th root of 3^n. And that equals just 3. So the limit is 3. That is the root limit, and that exceeds 1.
So therefore, by the root test I conclude that this thing must diverge, which of course, is old news since we knew it diverged before. This is confirmed by this. This thing diverges by the n^th root test. Here you can see the root test in action.
Let's try something a little bit more exciting, and where we don't know the answer in advance. So let's take the , so this is going to alternate a little bit, n + 1^n. Now what's going on here? Well, this is an alternating series. And what's happening with the denominator? This is a very fast growing denominator. I'm taking (n + 1)^n, so that's really growing fast, and it's growing faster than even a p-series we grow, because not only do I have a 2 in here, but I have a continually growing exponent. So my thinking is this is growing so fast to infinity, the whole thing must be shooting to zero at such a fast, rate this must converge. That's my guess. And I see n's raised to n powers. This is perfectly suited for the n^th root test. When you see n powers with n's here, the n^th root test is the way to go. So let's apply the n^th test and see what this says.
So I take the limit, it then goes to infinity of the n^th root of the absolute value of the n^th term. So what does that equal? That equals the limit as n goes to infinity of the n^th root of...now I've got to put in the divided by (n + 1)^n. What does the absolute value do? It annihilates every other negative sign that I have, so that drops out. The bottom is already positive, so I'm just left with the limit as n goes to infinity of the n^th root of . So what's that? Well, the n^th root of 1 is just 1, and the n^th root of (n + 1)^n - well, the n^th root and the n power kill each other off, and I'm just left with n + 1.
And what's that limit? Well, that limit is plainly zero, and so that is the root limit, which is plainly less than 1. It's as far from 1 as you can possibly get and still be positive. And so therefore, by the root test I can conclude that, in fact, this must converge. And in fact, it converges absolutely. So this thing converges absolutely by the root test.
Let's try one last one together. This one is the . Now I've got to be completely honest with you, because I've never lied to you. First thing I do when I look at this is I think two things. First of all, when I see n's here with an n power here, I immediately think of the root test. So in the bottom of my mind I'm going, "root test." So the next thing I do when I see this, I say, "Hey. What about the quickie test?" In fact, do the terms shrink to zero? Answer: no. If you take the limit and let n go to infinity, you have to use a little L'Hôpital's rule in a very clever way, but we've already seen that this thing is semi-familiar; this actually converges to the .
You can see that really fast if you want by remembering that we saw that this limit, limit as n goes to infinity . That actually works out to be e. If you work out that inside piece there, that's just , getting the common denominator, and this is the reciprocal of that. So this limit will actually be . So is not zero. These terms have to shrink to zero if there's any hope of this thing to converge. So already know that this diverges. But let's use the root test and see what happens.
We'll give the root test; take the limit as n goes to infinity of the n^th root of the absolute value of the n^th term. That's the limit as n goes to infinity of the n^th root of what? Well, the absolute value of this is just itself, so I just see . The n^th root of an n power, they kill each other off, so I'm going to put the limit as n goes to infinity of . What's that limit? One application of L'Hôpital's rule, it equals one. So the root limit is one. And so rho equals one, which means what? It means, whoop, test inclusive. We don't know whether this converges or diverges.
So lucky for us we had the foresight to even ask the quickie question, does this thing, in fact, shrink to zero? The answer is it absolutely does not, and so therefore, there's no hope of this converging. This diverges by the quickie test, because the terms themselves are not going to get small enough. They don't go to zero. They approach , and if you add up a lot of 's, that's going to be pretty big. So, in fact, here's an example where the n^th root test in fact is inconclusive, and yet we know the infinite series diverges, because the terms aren't getting small.
So that was the n^th root test and you can enjoy taking roots all over the place now.

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