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

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

We're going to employ the strategy of comparison.

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

b sub n, that is larger than or smaller

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

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

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

A test which has some complicated hypotheses.

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

Subsequent terms get smaller.

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

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

positive and decreasing and which agrees with

the sequence at every integer point. If we have

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

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

Of a of x, dx.

And these both have the same convergence or divergence.

If one

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

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

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

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

integrals can be tough. Its overall usefulness is limited.

But there are several key examples where it works well.

Why does this test work?

Well, we've seen it in use once

before, when we determined that the harmonic series.

Diverge.

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

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

series was bounded below by the improper integral.

If instead of a left remonsom we took a

right remonsom then we get a bound in the other

direction that the series or at least the tail,

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

Having this bound on both sides means that the series and

the integral have the same convergence or divergence properties.

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

Consider the absolutely fundamental series, sum 1 to

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

This should remind you of the p integrals from when

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

And the right way to understand their convergence

or divergence is to compare to the integral.

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

Dx.

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

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

to 1. Therefore, the same holds

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

You must memorize, a related example that is shown

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

to the pth power. If we continue this to

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

Then we can integrate that with respect to dx.

And obtain an integral using a little u substitution to get

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

Evaluating those, we see. Then again, we get

convergence when p is bigger than 1 and divergence when

p is less than or equal to 1.

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

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

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

test between two sequences. Our hypothesis are relatively simple.

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

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

If this is true,

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

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

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

But what can we say about convergence or divergence?

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

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

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

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

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

This is the integral test.

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

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

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

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

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

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

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

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

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

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

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

definitely bounded above by a sequence b sub n,

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

series, which definitely converges.

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

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

We could. Have bounded the a sub n by a

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

That p series converges and so the comparison test

would tell us that the original series converged as well.

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

1 over n. However, this p series

diverges and it doesn't tell us anything

about whether the a series converges or not.

This is the subtlety of using the comparison test.

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

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

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

We can bound the numerator.

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

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

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

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

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

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

converges. Therefore, the smaller a series converges

as well. Now let's see an example where

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

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

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

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

By square root

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

that is a p series with p equals 1.

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

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

So, what are we going to do?

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

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

Well let's let this be the b terms.

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

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

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

terms, since the a terms give us a p series

with p equals 1. The a series diverges and thus

does the b series diverge as well.

Now another example leads us to some complexity if

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

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

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

Well, we need to bound the numerator from above.

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

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

The denominator,

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

No.

N to the fourth?

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

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

Ugh, this gets very complicated.

Maybe, the comparison test is the wrong one.

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

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

Hello, class? Where did everyone go?

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

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

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

term is 2 over x squared.

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

This x goes to infinity.

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

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

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

Therefore, we know that this series converges

because we know the leading order term.

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

The limit test says, if you have 2 positive sequences

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

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

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

two series have the same convergence or divergence behavior.

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

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

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

Just as we did with improper integrals.

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

That looks very complicated to figure out,

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

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

But that something is very small.

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

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

And that's a p series, with p equals

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

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

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

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

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

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

When you get the exponents right, you see that the

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

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

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forget it.

That diverges and so does the original series.

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

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

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

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

Always look to compute the leading-order term.