College Calculus: Level II Comparison Test

Hi, we are checking out some more examples of the comparison test and the limit comparison test.0000

What we have right here is the sum of eⁿ/n!.0005

That one is a little tricky.0011

I want to write that one as eⁿ/n×1×2×3...×n.0013

What I notice here is this 3, 4, up to n, all of those numbers are bigger than or equal to 3.0031

You will see why the 3 is significant as a cut off in a second.0039

This is < or = eⁿ/1×2×3, and there were n-2 factors there.0044

So 3^n-2.0062

Just to make the algebra a little bit cleaner, I am going to multiply the top and bottom by 3².0064

3² × eⁿ/1 × 2, now I can say that is 3ⁿ.0071

I just multiplied in 3² × the top and bottom.0078

So that I can have an n in the exponent instead of an n-2.0082

This in turn is, 3² is 9/2 × eⁿ/3ⁿ.0085

This is (e/3)ⁿ.0094

That is the series that I am going to use as my b_n.0100

Remember a_n is always the series that you are given.0102

The point here is that a_n is < or = to b_n.0109

The sum of b_n converges.0114

Why does it converge?0118

Because, b_n is just a constant 9/2 × e/3ⁿ.0121

When you have a constant raised to the n power, it is a geometric series.0128

So you look at the common ratio, well here r = e/3.0136

Now e is about 2.7.0146

I do not know the exact value of e, but I know it is about 2.7.0149

The key thing there is that 2.7/3 is less than 1.0154

Remember 1 is the cutoff to check when a geometric series converges.0159

We have that e/3 < 1, so our common ratio < 1.0165

That means the sum of b_n converges.0171

So, our smaller series a_n, must also converge by the comparison test0175

Not the limit comparison, the original comparison test.0185

So, now it should be evident why I cut things off between 2 and 3 here.0193

I was kind of looking ahead to this common ratio.0202

I knew that e is about 2.7.0206

So, I saw that I had a bunch of numbers bigger than that in the denominator,0211

If I cut it off between 2 and 3.0218

That is why I cut off all of these number bigger than 3.0221

I used those to build the geometric series,0225

And then I just had to save the 1 and 2 separately.0229

1 × 2 just gave me a two in the denominator and did not really affect things later on.0232

The important thing was to save the e/3, and compare it with a geometric series.0238

Since we know the geometric series converges, and we have something smaller,0242

We can say our own series converges by the comparison test.0250

The last example that I want to do here is a little more complicated.0000

We have the series of n⁴+3n-8/3n⁷+6n⁵.0006

You see something complicated like this and it is important to focus on which terms are really going to affect it,0014

And which terms are not really going to play much role.0020

The key thing here is the two biggest terms, n⁴, in the numerator, and n⁷ in the denominator.0024

It looks like if you strip away the extraneous stuff there,0030

It looks like the sum of n⁴/n⁷.0038

That in turn would simplify down to 1/n³.0041

OK, I am going to call my 1/n³, that is going to be the b_n that I create.0051

This entire given series, that is a_n.0057

We will look at a_n vs. b_n,0062

We will divide them together and try to go for something using the limit comparison test.0065

a_n/b_n is n⁴+3n-8/3n⁷+6n⁵.0069

b_n is 1/n³,0081

But we can flip that up into the numerator so we can write that as n³/1.0084

That turns into n⁷+3n⁴-8n³/3n⁷+6n⁵.0094

Now you look at something like this and identify the biggest term anywhere,0107

Which is an n⁷ in the top and the bottom and you divide by it.0113

So, you get 1+3/n³-8/n⁴/3+6/n².0116

All of these other terms, go to 0.0134

This whole thing goes to 1/3.0140

The key thing about the 1/3 there, is only that it is a finite number, it is not infinity and it is not 0.0144

If you get a finite number there, the limit comparison test says that your given series does whatever the other series does.0152

It does the same thing that the other series does.0160

In this case, since the series we made ourselves, the b_n,0164

Well that is 1/n³, that is a p series and p is 3.0171

That converges.0176

The important thing about 3 there is that it is > 1.0187

Anything > 1 makes that converge.0190

Since that converges, we can say that the sum/a_n also converges.0192

Our justification there is the limit comparison test.0203

Either both series converge, or both series diverge.0208

Since the one we introduced converged, the given one converges as well.0211

To recap there, we are given some complicated series here.0217

We identify the important elements and use those important elements to build our own series that we introduce.0222

We then divide them together, take the limit,0230

If we get a finite non-0 limit,0234

Then we can say that both series do the same thing.0237

Hopefully the series that we introduced is a simple enough one where we can look at it and quickly say if it diverges or converges.0240

In this case it is a p series, and p is bigger than 1, so it converges.0248

So, we can say that the given series converges as well.0254

Thanks for watching, this is educator.com0259

Hi, this is Will Murray and we are here today to talk about the comparison test.0000

The way this works is that you will be given a series that we are going to call a_n,0006

And you create your own series that we will call b_n.0015

Then you will try to compare these 2 series to each other.0020

The way this works is, the one you create, if that one converges,0024

And the one that you are given is smaller than it,0029

Then you can say that the given series converges as well.0032

On the other hand, if the one you create diverges,0037

And the given series is bigger than the one you created,0040

Then you can say the given series diverges.0045

Now it is very important to get these inequalities going the right way.0048

If these inequalities are going the wrong way, then the comparison test does not tell you anything.0053

We will see some examples that illustrate the difference between those inequalities going the right way and going the wrong way.0057

We will be using a second test called the limit comparison test.0067

It starts out the same way,0069

You will be given a series, then you create your own series,0071

But then instead of checking the inequalities,0074

What you do is you divide those 2 series together and then take the limit as n goes to infinity.0077

If you get a finite and positive number, it cannot be 0, it has to be a positive number,0084

Then you can say whatever the series you created does, converges or diverges,0091

You can say that the given series does the same thing.0097

let us check that out with some examples.0101

The first one here is the series of 1/n+1.0105

That is the given series, the a_n.0109

When you look at this series, the simplest series that this seems to resemble,0113

Is the series 1/n.0123

Because 1/n+1, is about the same as 1/n.0126

So that is the series we create ourselves and we call the b_n.0131

Now let us try comparing those to each other.0134

1/n+1 works as 1/n, well n+1 has a bigger denominator.0138

Which means that 1/n is bigger.0146

So, 1/n is bigger, the b_n.0149

So, the a_n is less than the b_n.0155

The sum of b_n, that is 1/n, diverges.0162

We know that we saw that one before and that was the harmonic series.0170

Or, you can think of it as the p series, with p = 1.0178

We know that series diverges.0182

What we have is a series that is less than it.0185

If a series is less than a divergent series, that is not going the right way to use the comparison test.0189

So, comparison fails to give us an answer here.0197

Tell us nothing.0203

So, the comparison test does not tell us what this series does.0207

Instead, you have to try another test,0212

For example, you could try the integral test and that actually will give you a good answer.0215

You can check that out with the integral test.0223

It turns out that this series diverges,0226

But the important thing is that you cannot use the comparison test on this one.0228

Let us try it out with another one.0230

3ⁿ/2ⁿ - 1,0233

That looks like the sum of 3ⁿ/2ⁿ.0240

So a_n is the given series, for our b_n, we will use the sum of 3ⁿ/2ⁿ.0249

Then we will compare those to each other.0255

So, 3ⁿ/2ⁿ-1 vs. 3ⁿ/2ⁿ.0260

Well 2ⁿ-1 is a smaller denominator which means that 3ⁿ/2ⁿ-1 is actually a bigger number.0268

What that is saying is that a_n is bigger than b_n.0278

We know that the sum of the b_n's diverges.0285

The reason we know that is because it is a geometric series.0295

The common ratio is 3/2, because 3/n/2ⁿ is just 3/2ⁿ, and 3/2 is bigger than 1.0300

So that is a geometric series that diverges,0309

And here we have a bigger series, so this bigger series, we can say it diverges by the comparison test.0313

In this case, the inequality did go the right way.0329

So, we get a conclusion by the comparison test.0333

Just a recap there, we looked at the series we were given, we tried to find the series that was similar to it,0338

And simple enough that we could answer pretty quickly whether it converged or diverged.0346

In this case it was a geometric series.0350

Then we compared them to each other.0354

The inequality does go the right way, so we can make this conclusion.0356

Next example I wanted to try is the sum of sqrt(2n+17)/n.0361

Now, this one most closely looks like,0369

Well, the sqrt(2n+17) that is really more or less the sqrt(n).0376

sqrt(n)/n, and so that in turn is sqrt(n)/n 1/sqrt(n).0385

That is the series we are going to use as our b_n.0398

This series is our a_n.0401

What we are going to do is divide those together because we are going to try to use the limit comparison test this time.0404

So we look at a_N/b_n,0411

And that is the sqrt(2n+17)/n.0414

All of that divided by b_n, which is 1/sqrt(n).0420

We can flip that fraction in the denominator up the numerator,0424

So we get sqrt(2n+17).0429

The sqrt(2n+17) × sqrt(n)/n,0435

Which is sqrt(2n²) + 17n/n.0444

If we look at the top and bottom there, the biggest terms we have in the top, we have an n² with a square root over it.0451

That is essentially n and in the bottom we have n.0457

So we would divide top and bottom by n, and we get,0463

If we bring that top end under the square root, we get 2+17/n.0467

Because when the n comes under the square root it turns into an n².0472

Then just 1 in the denominator.0477

17/n goes to 0.0480

So, the whole thing goes to sqrt(2).0483

The important thing about the sqrt(2) is that it is a finite number, it is not infinity.0488

It is not 0, so the limit comparison test applies.0493

Since the sum of b_n diverges, well how do we know that,0498

That is because we can think of 1/sqrt(n) as 1/n^1/2.0509

That is a p series, with p equal to 1/2.0515

The key thing there is that 1/2 < or = 1.0522

That means that b_n is a divergent series.0527

The limit comparison test says that the given series a_n does the same thing that your series b_n does.0530

So, we can conclude that the sum of a_n also diverges by the limit comparison test.0539

Again, the key point there is that we look at the series that we are given, and we try to find,0550

So that is the given series,0562

We try to find the given series that behaves like the series that we are given but is simpler to deal with.0565

What I did here was essentially strip away the 2 and the 17,0569

Because those were not so important, and then I called the new series b_n.0574

Then we try to compare those series to each other.0578

This time by dividing.0582

If we get this finite non-zero number at the end of it,0584

That says that both series behave the same way.0588

Since we know the b_n diverges, because it is a p series,0593

Then we can say the a_n diverges as well and justify that conclusion using the limit comparison test.0597

So, another example I would like to look at is the sum of sin(1/n).0605

The way we might think about that is,0610

let us think about the graph of sin(x).0614

As n goes to infinity, 1/n goes to 0.0618

So let us look at the graph of sin(x) when x is very near 0.0625

What we see is that the graph of sin(x),0629

This is sin(x), is very close to the graph of x as x approaches 0.0637

Sin(x) is almost the same as x.0647

Sin(1/n) the a_n might behave like the series b_n = 1/n.0650

Because sin(x) behaves like x.0660

Let us try that out.0662

We will take the series b_n = 1/n, and then we will look at a_n/b_n.0665

Is sin(1/n)/1/n.0673

Remember as n goes to infinity, 1/n goes to 0.0679

So this is sin(0) which goes to 0.0684

1/n also goes to 0, so we have a 0/0 situation.0686

That is a situation where you can use l'Hopital's rule.0690

So l'Hopital's rule says you can take the derivatives of the top and bottom,0695

I will take the derivative of the bottom first because it is easier.0703

The derivative of 1/n is just 1/-n².0707

The derivative of sin(1/n) is cos(1/n).0710

× derivative of 1/n by the chain rule which is -1/n².0718

Those -1/n²'s cancel.0724

We get cos(1/n) and if we take the limit of that,0727

As n goes to infinity, that is cos(0) which is 1.0735

The key thing about 1 here is only that it is a finite non-zero number.0742

If it is a finite non-zero number, that says whatever one series does, the other series does.0752

So, let us ask ourselves,0760

The sum of b_n does what?0763

Well b_n is just 1/n, and we know that is the harmonic series.0765

And it is also a p series.0771

So, either one of those is just vacationed since we already showed that it diverges.0775

It is a p series with p = 1.0780

So that is a divergent series.0788

Since that series diverges,0792

We can say that the given series a_n also diverges by the limit comparison test.0795

So that one was a little bit trickier, probably was not so obvious what series we should compare it too.0810

The key thing there was realizing that sin(1/n) is a lot like 1/n when n goes to infinity.0816

Because 1/n was going to 0.0825

sin(x) was very similar to x.0828

Once we figure out which series we want to compare it to, we divide them together, take the limit,0832

Which uses l'Hopital's rule, we get a finite non-zero number,0838

And that says the two series do the same thing.0843

Since one of them diverges, we can say that the given series diverges as well.0847