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>> Our goal in this lesson is to determine convergence or

Â divergence of a series based on a sequence as of N.

Â We're going to employ the strategy of comparison.

Â That is, we'll work with some other sequence, say,

Â b sub n, that is larger than or smaller

Â than a sub n. Or instead of another sequence we might

Â use a continuous function that is related to the original sequence.

Â This is the basis of our first test, the integral test.

Â A test which has some complicated hypotheses.

Â First of all, the sequence a sub n, must be positive and decreasing.

Â Subsequent terms get smaller.

Â Second, we need to take that sequence and connect the dots, turning

Â it in, to a continuous function, a of x. Is it itself

Â positive and decreasing and which agrees with

Â the sequence at every integer point. If we have

Â both of these, then we can compare the series, the

Â sum of the a sub n's with the integral if x goes from 1 to infinity.

Â Of a of x, dx.

Â And these both have the same convergence or divergence.

Â If one

Â converges, so does the other. If one diverges, so does

Â the other. Now, in terms of applicability, this test

Â is not stellar. It has some strict hypotheses, and in

Â terms of ease of use, this is also not the simplest test since improper

Â integrals can be tough. Its overall usefulness is limited.

Â But there are several key examples where it works well.

Â Why does this test work?

Â Well, we've seen it in use once

Â before, when we determined that the harmonic series.

Â Diverge.

Â We compared it to the continuous inter-grand /n 1 over x.

Â We saw, by taking a left riemann sum, that the

Â series was bounded below by the improper integral.

Â If instead of a left remonsom we took a

Â right remonsom then we get a bound in the other

Â direction that the series or at least the tail,

Â ignoring the first term, bounds from below the improper integral.

Â Having this bound on both sides means that the series and

Â the integral have the same convergence or divergence properties.

Â If we look at a few examples, we'll see better how to apply this.

Â Consider the absolutely fundamental series, sum 1 to

Â infinity, of 1 over n to the p, where p is a constant.

Â This should remind you of the p integrals from when

Â we did improper integrals. In fact these are called the p series.

Â And the right way to understand their convergence

Â or divergence is to compare to the integral.

Â As x goes from one to infinity of one over x to the p.

Â Dx.

Â Now we know, that these p integrals converge when p is strictly greater

Â than 1. And diverge when p is less than or equal

Â to 1. Therefore, the same holds

Â for the p series. This is a set of examples.

Â You must memorize, a related example that is shown

Â in the same manner. Is the sum of 1 over n times log of n

Â to the pth power. If we continue this to

Â the function. 1 over x times log of x to the p.

Â Then we can integrate that with respect to dx.

Â And obtain an integral using a little u substitution to get

Â log of x to the 1 minus p over 1 minus p.

Â Evaluating those, we see. Then again, we get

Â convergence when p is bigger than 1 and divergence when

Â p is less than or equal to 1.

Â Thus, the same convergence and divergence holds for this series.

Â And in both these cases, we have a range of convergence for p.

Â Where it's strictly greater than one. Our next test is a comparison

Â test between two sequences. Our hypothesis are relatively simple.

Â We have two sequences, a sub n and b sub n.

Â Both are positive, and the b sub n's are bigger than the a sub n's.

Â If this is true,

Â then what can you say about the series? Well, if b sub n is bigger than or

Â equal to a sub n, then the sum of the b sub n's is bigger than

Â or equal to the sum of a sub n's. That much is clear.

Â But what can we say about convergence or divergence?

Â Well, because all the terms are positive, if the larger of the

Â two series converges, then the smaller one must converge as well.

Â Now, the converse of this is not. Necessarily, so but the

Â [UNKNOWN]

Â positive is if the smaller series diverges, then so does the larger.

Â Now in terms of applicability oh, this one doesn't have so many hypothesis.

Â This is the integral test.

Â But in terms of ease of use and usefulness, well it's not perfect.

Â In part, because it's easy to get these two results

Â confused, and it's sometimes difficult to pick the appropriate

Â b sub n, or a sub n, or to know which is which.

Â Let's see this in the context of some examples.

Â If we look at the sum, as n goes from 0 to infinity, of 1

Â over quantity, 4 plus negative 1 to the n, to the n.

Â That's 1 plus a third plus 1 over 5 squared, plus 1

Â over 3 cubed, plus 1 over 5 to the fourth and etcetera.

Â That's a funny looking series but it's not that hard to

Â figure out because if we called this terms the a sub n's, they are

Â definitely bounded above by a sequence b sub n,

Â given by one of our 3 to the n. Summing that up gives us a geometric

Â series, which definitely converges.

Â Therefore, the a series converges as well. Now, one of the problems

Â is no one told us which is a, which is b, and what we should use.

Â We could. Have bounded the a sub n by a

Â p series. Where p in this case is equal to 2.

Â That p series converges and so the comparison test

Â would tell us that the original series converged as well.

Â On the other hand, a perfectly good upper bound is given by

Â 1 over n. However, this p series

Â diverges and it doesn't tell us anything

Â about whether the a series converges or not.

Â This is the subtlety of using the comparison test.

Â P series tend to be extremely helpful when using the comparison test.

Â Consider the sum from 0 to infinity of cosine squared n over 1

Â plus the square root of n cubed. Let's let that be our a sub n.

Â We can bound the numerator.

Â From above by one since cosine squared is always between 0 and 1.

Â What can we do with that denominator? Well, the denominator is strictly larger

Â than n to the 3 halves. Therefore, we can bound

Â this a sequence by a b sequence of the form, 1

Â over n to the 3 halves. Since we know something about the p

Â series, for p equals 3 halves, we know that the larger series

Â converges. Therefore, the smaller a series converges

Â as well. Now let's see an example where

Â it runs in the other direction. Consider the sum of our n of 1

Â plus square root of n. Over square root of 1 plus n cubed.

Â Now if we let that term be our a sub n, then we can bound the numerator

Â by, let's say, 2 square root of n. And the denominator.

Â By square root

Â of n cubed. Now, that's going to give us a b sub n

Â that is a p series with p equals 1.

Â That is not going to help us. That series diverges.

Â It doesn't tell us anything about the a series.

Â So, what are we going to do?

Â Well, we implicitly assumed that our given was the a.

Â But there's no reason why that has to be the case.

Â Well let's let this be the b terms.

Â And we can choose an a term that bounds from below.

Â Taking a numerator of the square root of n, and a denominator of let's say

Â 2n cubed. Then we get a lower bound for the b

Â terms, since the a terms give us a p series

Â with p equals 1. The a series diverges and thus

Â does the b series diverge as well.

Â Now another example leads us to some complexity if

Â we inlet a sub n b to n cube plus 3 n minus

Â 8 over n to the fifth minus 5 n cubed minus n squared plus 2.

Â And what are we going to choose, when we try to do the comparison test?

Â Well, we need to bound the numerator from above.

Â Let's say 2 n cubed, no that's not going to work, because we've got the 3 n.

Â Let's say 3 n cubed. That will do the job.

Â The denominator,

Â well we can bound from below by 2n to the fifth?

Â No.

Â N to the fourth?

Â No, that's not going to be good, that gives us

Â a harmonic series for the v sub n, that diverges.

Â Ugh, this gets very complicated.

Â Maybe, the comparison test is the wrong one.

Â Instead, let's try an integral test. Who wants to integrate 2x cubed plus

Â 3x minus 8 over x to the 5th minus 5x cubed minus x squared plus 2.

Â Hello, class? Where did everyone go?

Â Well, what are we going to do in this case?

Â Asymptotic analysis is going to save us as it has done so many times before.

Â Taking this, a of x we see that the leading order

Â term is 2 over x squared.

Â Everything else is in big O of 1 over x to the 4th.

Â This x goes to infinity.

Â Well the same thing holds for the discrete setting with our sequence

Â a sub n. Since the leading order term, is a p

Â series with p plus 2, we know that that converges.

Â Therefore, we know that this series converges

Â because we know the leading order term.

Â 13:19

That's the idea behind our last test in this lesson, the limit test.

Â The limit test says, if you have 2 positive sequences

Â that have the same asymptotics, the same meeting order term/g.

Â Specifically, if their limit, of the quotient is n goes to infinity,

Â is some number that is strictly between 0 and infinity, then these

Â two series have the same convergence or divergence behavior.

Â This is a wonderful test, very easy to apply and to use.

Â And it's overall usefulness is very high as well.

Â Let's see some examples where we compute the leading order terms.

Â Just as we did with improper integrals.

Â Consider the sum over n of log squared of quantity 1 plus 1 over n squared.

Â That looks very complicated to figure out,

Â I don't want to try an integral test with that, I don't know what to compare it to.

Â But I do know the Taylor series for log of 1 plus something.

Â But that something is very small.

Â I know that the leading order term for this seqeunce is 1

Â over n squared quantity squared. That is 1 over n to the 4th.

Â And that's a p series, with p equals

Â 4, it converges. Thus, does my original series converge?

Â Here's another crazy looking example, we'll put a polynomial on

Â top, and something with some square roots in it down below.

Â All we need to do, is compute the leading order term for the numerator,

Â it's an n squared with a 3 in front but forget about the constant.

Â In the denominator, the leading order term, is n times n to the 5 3rds.

Â When you get the exponents right, you see that the

Â leading order term is 1 over m to the 2 3rds.

Â That's a p series with p equals 2 3rds.

Â [SOUND]

Â forget it.

Â That diverges and so does the original series.

Â Lastly, consider n times pi to the n over the hyperbolic cosine of n.

Â Well the leading order term in hyperbolic cosine is exponential in n.

Â Therefore, with a little bit of factoring, we see that the leading-order term

Â is n, times pi over e, to the n. This definitely diverges.

Â Always look to compute the leading-order term.

Â