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Lecture Comments (12)

1 answer

Last reply by: Dr. William Murray
Thu Apr 24, 2014 6:13 PM

Post by Jack Miars on April 19, 2014

In lecture example three, when solving for s_n you put that the terms that don't get cancelled are 1/n+1 and 1/n+2, however if you look at it for when n=3, the values are 5 (n+2) and 6 (n+3).  Is this an error or am I mistaken?

3 answers

Last reply by: Dr. William Murray
Wed Nov 20, 2013 1:41 PM

Post by Narin Gopaul on October 22, 2013

Good day DR. I just took my second test in calculus and failed miserably. I used your videos which did help me to get the theory down. The problem am facing is that the (Functions) you see on a test where you will apply the theory on to solve is extremely hard. How do I solve this problem when there is no telling of what type of functions could appear.

I will appreciate any advice you can give!!!

1 answer

Last reply by: Dr. William Murray
Sun Sep 22, 2013 8:38 AM

Post by Mark Bogenrieder on September 17, 2013

Looking at this particular problem, what exactly clues you in to use partial fraction decomposition? You go right to PFD without really explaining what led you to that solution. Thanks.

1 answer

Last reply by: Dr. William Murray
Thu Apr 25, 2013 2:03 PM

Post by vishal patel on April 24, 2012

Wish I had Professor Murray for my course.

1 answer

Last reply by: Dr. William Murray
Thu Apr 25, 2013 2:02 PM

Post by Luis Robles on March 27, 2012

Great explanation!
Keep up the good work!

Series

Main definitions and theorem:

Definitions:

  • The notation stands for the series a0 + a1 + a2 + …

  • The sequence of partial sums of the series is the sequence

  • The phrase the series an converges (to a limit L) means that the sequence of partial sums {sn} converges (to L).

  • The same definition holds for diverges, diverges to ∞, and diverges to −∞.

A geometric series is one where each term is the previous term multiplied by the same common ratio:

Theorem (Test For Divergence): If is a series, and the sequence {an} converges to something other than 0, or if it diverges, then the series diverges.

Hints and tips:

  • Remember to distinguish between the sequence of terms {an}, the series an, and the sequence of partial sums {sn}. All three are different!

  • In determining whether a series converges or diverges, you can ignore the first few terms. What is important is what the later terms do. However, the first few terms do matter in determining what limit a series converges to.

  • Don’t try to analyze series by plugging numbers into a calculator. This is extremely unreliable.

  • The Test For Divergence is usually the first and easiest test to check for series. However, remember that it is a one way test. If the sequence does converge to 0, then TFD tells you nothing about the series.

  • You can sometimes use partial fractions to analyze a series involving rational functions. Then write out some partial sums of the series and see if the terms cancel. These are called telescoping series.

  • It is sometimes useful to invoke algebraic identities such as ln a/b = ln a − ln b.

  • There are several different versions of the formula for the sum of a geometric series, depending on whether the series starts at n = 0, n = 1, and so on. The easiest one to remember, which works in all situations, is

    	   first term   
    	1 − common ratio

    assuming the ratio has absolute value less than 1.

Series

Does ∑n = 1 7(2)n − 1 converge or diverge?
Determine rr = 2|2| > 1, thus the series diverges
Does ∑n = 1 [(3n)/3] converge or diverge?
  • Simplify term
  • n = 1 [(3n)/3] = ∑n = 1 3n − 1
Determine rr = 3|3| > 1, thus the series diverges
Does ∑n = 1 18( [1/3] )n − 1 converge or diverge?
Determine rr = [1/3]|[1/3]| ≤ 1, thus the series converges
What is the sum of the first 9 terms of series an = 7(2)n − 1
  • Use the Sum of the first n terms equation
  • Sn = [(a(1 − rn))/((1 − r))]
  • Define variables
  • a = 7
  • r = 2
  • n = 9
  • Apply equation
  • Sn = [(a(1 − rn))/((1 − r))]
  • = [(7(1 − 29))/((1 − 2))]
= 3577
What is the sum of the first 3 terms of series an = 18( [1/3] )n − 1
  • Use the Sum of the first n terms equation
  • Sn = [(a(1 − rn))/((1 − r))]
  • Define variables
  • a = 18
  • r = [1/3]
  • n = 3
Sn = [(a(1 − rn))/((1 − r))] = [(18( 1 − ( [1/3] )3 ))/(( 1 − [1/3] ))] = 26
What is the series represented by 1 + [4/5] + [16/25] + [64/125] + ...
Change terms into the contest of squares
1 + [4/5] + [16/25] + [64/125] + ... = 1 + [4/5] + [(42)/(52)] + [(43)/(53)]an = ( [4/5] )n − 1
Does the series, 1 + [4/5] + [16/25] + [64/125] + ..., converge?
Determine rr = [4/5]|[4/5]| < 1, thus the series converges
What is the sum of the first 5 terms of the series 1 + [4/5] + [16/25] + [64/125] + ...
  • Use the Sum of the first n terms equation
  • Sn = [(a(1 − rn))/((1 − r))]
  • Define variables
  • a = 1
  • r = [4/5]
  • n = 5
Sn = [(a(1 − rn))/((1 − r))] = [(( 1 − ( [4/5] )5 ))/(( 1 − [4/5] ))] = 3.36
What is the sum of the series 1 + [4/5] + [16/25] + [64/125] + ...
  • Use the Sum of series equation
  • S = [a/(1 − r)]
  • Define variables
  • a = 1
  • r = [4/5]
S = [a/(1 − r)] = [1/(1 − [4/5])] = 5
What is the sum of the series ∑n = 1 3( [3/2] )1 − n
  • Alter the sequence with exponent properties
  • n = 1 3( [3/2] )1 − n = ∑n = 1 3( [2/3] )n − 1
  • Use the Sum of series equation
  • S = [a/(1 − r)]
  • Define variables
  • a = 3
  • r = [2/3]
  • n = 5
  • Apply equation
  • S = [a/(1 − r)]
  • = [3/(1 − [2/3])]
= 9

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Series

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Important Definitions 0:05
    • Sigma Notation
    • Sequence of Partial Sums
    • Converging to a Limit
    • Diverging to Infinite
  • Geometric Series 2:40
    • Common Ratio
    • Sum of a Geometric Series
  • Test for Divergence 5:11
    • Not for Convergence
  • Lecture Example 1 8:32
  • Lecture Example 2 10:25
  • Lecture Example 3 16:26
  • Additional Example 4
  • Additional Example 5

Transcription: Series

Hi, this is educator.com, and we are trying some examples of finding the limits of series.0000

Our example right now is sum of ln(n+1/n).0008

let us take a look at that.0014

The ln(n=1/n), we can invoke some rules of natural logs there.0016

We can write that as ln(n+1)-ln(n).0023

That will be very useful when we try to determine whether this series converges or diverges,0029

And what it converges or diverges to.0034

Remember, the way you determine whether a series converges or diverges is you look at the partial sums.0038

You write those out as a sequences,0044

Then you look at whether that sequence converges or diverges and where it goes.0048

Let us write out some partial sums here.0052

The first partial sum, s1, is just the first term here.0057

So, ln(2) - ln(1).0064

We could simplify that, of course ln(1) = 0,0068

But I think it will be easier to spot a pattern, if we do not simplify that.0072

let me go ahead and find s1 is just the s1 term, ln(2) - ln(1),0077

+ the n = 2 term, ln(3) - ln(2).0084

s3, ln(2) - ln(1) + ln(3) - ln(2) + ln(4) - ln(3).0092

We start to see some cancellation here.0109

ln(2) cancels, ln(3) cancels, and then we are just left with the beginning and ending term.0112

Let us try one more, s4.0118

Is ln(2) - ln(1) + ln(3) - ln(2) + ln(4) - ln(3) + ln(5) - ln(4).0120

Again, we have lots of cancellation.0136

ln(2), ln(3), ln(4), and we are left with beginning and ending terms.0138

A general formula for the nth partial sum would be -ln(1) +,0147

When we had n = 4, we had ln(5), so the general pattern would be ln(n+1).0159

Then that we could simplify down.0164

ln(1) is jsut 0, so that is just ln(n+1).0166

That is what we get when we look at the sequence of partial sums.0173

We want to ask what happens when n goes to infinity.0179

Take the limit of that as n goes to infinity.0181

As n goes to infinity, we are plugging in bigger and bigger values to natural log.0184

Natural log goes to infinity itself.0190

So the sequence of partial sums diverges to infinity.0195

So the series by definition,0200

We say that diverges, it does the same thing that the sequence of partial sums does.0212

It diverges to infinity as well.0220

Let us try one more example of a series problem.0000

We are given the summation of n=2 to infinity of 3n/43n+2.0004

Again, I want to try to investigate this by writing out a few terms and seeing what happens.0013

If we plug in n=2, we get 32/43×2 + 2.0019

So that is 48.0028

Now n=3 gives us 33, over 43×3+2, so that is 411,0031

+ 34/43n+2, so 414.0041

What you might notice here is that each one of these terms is a common ratio multiplied by the previous term.0056

This second term = the first term × 3/43.0064

To get from the second to the third term, we multiply by 3/43.0073

So, each one of these terms, you get it by multiplying the previous term by 3/43.0082

So, what we have here is a geometric series.0090

Our common ratio is r = 3/43.0097

Which is certainly less than 1.0107

The reason I bring that up is because we want to check whether the geometric series converges,0110

You have to check whether the absolute value of 3 < 1.0118

And, certainly 3/43 < 1.0122

So, it converges and we have a formula for what a geometric series converges to. 0125

Remember that I said the easy way to remember that formula,0135

Is the first term/1-the common ratio.0142

That is the sum of a geometric series.0161

i think that is easier and more reliable than any numerical formula you can get.0164

Here, our first term is 32/48.0170

The common ratio is 3/43.0176

So, that is a little bit messy.0186

We can clean it up a little bit by multiplying top and bottom by 43.0189

That will give us 32/45.0195

On 43 - 3 in the denominator.0199

That is 9/45/64-3.0205

So, that in turn becomes 9/61×45.0214

Again, the principle of dealing with this series is to write out the first few terms.0226

Recognize that it is a geometric series.0235

Recognize that each term is the previous term multiplied by a common ratio.0238

Identify the common ratio, see if it less than 1, and if it is, you can say right away that the series converges.0244

Then you can invoke this formula, first term divided by 1-common ratio.0252

Do a little bit of simplification, and find the sum of the series.0259

This has been Will Murray for educator.com.0265

Hi this is educator.com0000

We are going to talk today about series0003

There are several bits of notations and definitions before we look at some examples0005

The first is this big sigma notation.0014

The notation this symbol sigma stands for the series where you plug in different values of n.0017

This is just short hand a0, a0 + 1, a0 + 2.0025

The way you want to think about these series is by thinking about the sequence of partial sums.0030

What you do is you add up these series 1 term at a time.0039

First you start with the 0 term if there is one. 0041

Then the second partial sum is a0 + a1,the third is a0 + a1 + a2 and so on.0047

The sn is a0 + a1 up to an0054

You call this the sequence of partial sums.0060

s0, s1, s2 up to sn.0064

You think of that as a sequence, not a series.0069

This is a sequence sn.0075

We want to say what it means for a series to converge.0081

You have to be quite careful when you make this definition.0091

It is not quite as obvious as you think.0094

What you say is you look at the sequence of partial sums.0097

We view that as a sequence0102

Then we go back to our definition for a sequence converging.0105

if the sequence of partial sums converges, to a particular limit.0111

Then we say the series converges to that limit.0117

By definition, a series converging means the sequence of partial sums converges.0122

In several of the examples that we will see later on,0127

We will be given a series and the way we will handle it is we will write out the partial sums and then we will think of them as the sequence.0130

We will see what happens to that sequence.0136

The same thing holds for all the other possibilities.0141

Diverging, diverging to infinity, or diverging to negative infinity.0145

You look at what the sequence of partial sums does.0151

Whatever behavior that does, you say the series does the same thing.0155

There is a very common type of thing that you will see.0162

We call it a geometric series.0165

It is one where each term is equal to the previous term multiplied by the same common ratio.0167

In practice that looks like a + some a × r + a × r2 and so on.0175

The important thing is that each term is getting multiplied by the same number every time.0182

We have a formula for the sum of a geometric series.0189

What you have to do is determine first of all if the ratio and absolute value < 1 or > 1 or = 1.0193

If the ratio and absolute value < 1, the series adds up to a/1-r.0204

This is the formula you will see in a lot of books.0212

I think this formula can be a bit misleading.0214

Because it depends on whether you start the series at n = 0, or at n = 1.0217

Some books have the a/1-r formula, some books have a slightly different formula.0226

It depends on whether they are using n = 1 or n = 0 as the first term in the series.0234

I do not like that formula so, I will give you a fail safe formula that will work in all situations right here.0240

First term, divided by 1 - the common ratio of the series.0246

That formula always works in all geometric series.0252

When your ratio is < 1.0255

I think that is the one to remember, even though it is words instead of an equation.0260

That is the one that will get you through any geometric series.0265

If the common ratio is bigger than 1 in absolute value0270

Or if it is equal to -1, than the series just diverges.0274

If the common ratio, if r = 1, then that means that what you are doing is you are adding a + a + a.0280

That clearly diverges either to infinity, if a > 0 , or -infinity, if a is negative.0292

You will probably not get geometric series with r - 0 because they are too simple to be given in a calculus exercise. 0303

One more theorem that we are going to be using is called the test for divergence.0311

It is usually the first thing that you want to test with every series.0318

If you are given a series, what you do is you look at the individual terms.0320

You see whether the individual terms converge.0326

If they converge you ask what do they converge to.0332

If they converge to anything other than 0, then you can immediately say the series diverges.0336

Also, if the sequence of terms diverges, you can say the series diverges.0347

To recap, you look at the individual terms, if they converge to something other than 0, or if they diverge.0354

Immediately you can say the series diverges.0360

The flip side that often mixes up students.0366

If the sequence of terms does converge to 0, people think that you can use that to conclude that the series converges.0369

That is not true and we will see examples of that later on.0391

Then just from that information you cannot conclude anything about the series.0398

You have to go and find one of the other tests that we will discuss later on.0416

If the sequence converges to 0, you are really stuck.0421

You cannot use the test for divergence, however, if the sequence converges to something other than 0,0425

You can use the test for divergence immediately to say that the series diverges.0430

To emphasize here, the test for divergence can tell you that a series diverges, but it can never tell you that a series converges.0439

This is a common mistake made by students.0482

People will say oh a series converges by the test for divergence.0488

That is a very bad mis-use and your teachers will have no patience with that.0491

You can use the test for divergence to say that a series diverges,0498

But if you get that the series converges, if the sequence converges to 0, the test for divergence tells you nothing and you have to find something else.0500

Let us try some examples.0510

First example here is the series n-1/n0513

Right away we will look at the test for divergence and see if it works. 0516

an = n-1/n.0520

We rewrite that as 1-1/n.0525

The limit of the sequence an is well the 1/n will go to 0 is 1.0531

This is not 0.0541

So the sequence of terms converges to something other than 0.0545

The sequence an converges to something other than 0.0560

The series diverges by the test for divergence.0580

The test for divergence says you look at the sequence of terms, see if they converge to something other than 0.0599

If so, then the whole series diverges.0608

If this had come out to be 0, if the limit had been 0, then we would not know and we would have to go on and find some other test there.0614

Let us try another example here.0626

The series of 1/n so again you will look at the test for divergence.0627

an = 1/n and the limit of that as n goes to infinity is 0.0637

The sequence converges to 0. 0645

The test for divergence, if the sequence converges to 0, tells us nothing.0649

So, t(d) fails to give us an answer here.0656

We cannot say anything yet.0663

Instead we will look at the partial sums.0664

s1 is just the first partial sum, that is just 1. 0676

s2 is 1 + 1/2.0679

s3 is 1 + 1/2 + 1/3.0684

s4 is 1 + 1/2 + 1/3 + 1/4.0691

I am going to start adding these numbers up and see what kind of sums we get.0703

Just look at 1, it sounds kind of silly to say it, but 1 > or = 1.0708

We will see why I am making a big deal out of that later on. 0715

1 + 1/2 > or = 3/2, in fact it is equal to 3/2.0721

I will skip s3, I will not look at that.0726

s4, if you look at 1/3 + 1/4 now 1/3 is bigger than 1/4.0730

So this is > or = to 1/4 + 1/4.0737

What we have here is 1 + 1/2.0743

Plus something bigger than 1/2.0747

This is bigger than 2.0750

Now I am going to start skipping, I am going to look at sn.0756

sn is 1 + 1/2 + 1/3 + 1/4 + 1/5 all the way up to 1/8.0760

Again, we have 1, that is bigger than 1.0774

1/2 is at least 1/2.0777

1/3 + 1/4 is bigger than or equal to 1/2.0783

Here we have 4 terms, each one of those is bigger than or equal to 1/8.0788

Collectively they are bigger than or equal to 1/2.0794

What we have here is 1 + 1/2 + 1/2 + 1/2.0802

This whole thing is < or = to 5/2.0807

Now I will skip to s16.0812

Without showing all the individual terms, it is bigger than 1/2 + 1/2 + 1/2 + 1/2 and then we will have 8 more terms, all bigger than 1/16.0816

So 1/2, this is bigger than or equal to 3.0829

If you look at the sequence of partial sums as a sequence, we have got 1, 3/2, 2, 5/2, 3.0834

Clearly, the sequence of partial sums sn diverges to infinity. 0845

By definition, the series 1/n diverges to positive infinity also.0868

There are a couple of points we want to make to recap that.0884

We tried to use the test for divergence.0886

That says you use the sequence for individual terms, but those went to 0, so the test for divergence tells us nothing.0891

Instead, we look at the sequence of partial sums.0897

That means we start adding up these terms and we kind of group them together in clever ways.0900

When we group them together in clever ways, we notice that the sequence of partial sums is going to infinity.0912

So, we say the entire series diverges to infinity.0921

A couple more things that I want to say about this series.0923

One is that it is very well known, it is called the harmonic series.0927

It is called the harmonic series because it arises in looking at musical notes.0934

This is well known and is called the harmonic series. 0941

People will tell you the harmonic series diverges and the reason is by this proof here.0945

Another important thing to note about this series is we could write this as the sum of 1-np where p=1.0952

It is really 1/n1.0960

This is a special case of something we are going to see later called the p series.0963

That is kind of a preview of something we will see later when we talk about the integral test.0972

We will be talking about p series.0976

The harmonic series is a special case of p series, with p=1.0978

Let us try another example here.0986

We are trying to determine if the series 1/n+1 converges or diverges.0990

Again, the trick here is to look at the partial sums and before we right out the sequence of partial sums.0999

We are going to do a little algebra here.1012

We are going to try to use partial fractions on 1/n × n+2.1016

Partial sums is an algebraic trick that we learned back in the partial sums section.1023

It says we can separate 1/n × n+2.1029

It says we can separate a/n + b/(n+2).1034

Then we can try and solve for constants a and b.1041

We learned this in detail in the lecture on partial fractions.1046

If you are a little fuzzy on partial fractions, we might want to go back and review that lecture,1050

In the meantime, I am not going to work it out but I am going to tell you the answer is a=1/2 and b=-1/2.1057

That comes from applying partial fractions here.1069

There is a little bit of algebra and solving 2 equations and 2 unknowns that I am suppressing here.1073

But, you can work it out and get these values for a and b.1080

We can write this as 1/2/n - 1/2/n+2.1085

If I factor out the 1/2 there, I get 1/2 × 1/n - 1/n+2.1092

That is going to be very useful in trying to figure out what the partial sums are.1102

Let me write down what a few of those partial sums are now.1106

s1, plug n=1 in here and we get 1/2 × 1 - n=1, gives us 1/3.1108

I could simplify that but I am not going to because later on it will be easier to spot a pattern if we do not simplify that.1119

s2 is 1/2 × 1 - 1/3, + the n = 2 term, is 1/2 - 1/4.1127

s3 is 1/2 × 1 - 1/2 + 1/2 × 1/2 - 1/4 + 1/2 × the third term is 1/3 - 1/5.1140

Now you start to notice some cancellation here.1160

This 1/3 will cancel with that 1/3, which is why I did not want to simplify earlier.1164

Let me write out 1 more term, s4.1167

1/2 × 1/3 + 1/2 × 1/2 - 1/4 + 1/2 × 1/3 - 1/5 + 1/2 × 1/4 - 1/6.1172

Now you start to see a lot of cancellation.1191

This 1/3 cancels with this 1/3, this 1/4 cancels with this 1/4. 1196

If there were another term, the next term would have a 1/5, and that would cancel with that 1/5.1200

So, it looks like there is going to be a lot of cancellation as we write out more partial sums. 1205

Let me try to write a general term for sn.1216

It is going to be 1/2, well after we cancel everything, the only terms left are this 11218

That never cancels, the 1/2 never cancels.1226

Then the very last terms do not cancel, but everything in the middle is all cancelled.1230

We get 1/2 × 1 + 1/2 -, well the last two terms.1240

When n=4, the last 2 terms were 1/5 and 1/6.1249

Those last two terms are n+1 and 1/n+2.1255

So, this simplifies down to 1/2 × 3/2 - 1/n+1 -1/n+2.1262

The limit as n goes to infinity of the partial sums.1279

Remember we think of the partial sums as a sequence now.1284

Those terms just go to 0.1289

We are left with 1/2 × 3/2 = 3/4.1292

We have the sequence of partial sums going to 3/4.1298

We say the series converges to 3/4.1304

To recap there, when you are given a series,1325

You want to determine whether it converges or diverges.1331

That depends on what the sequence of partial sums does.1336

That is s1, s2, s3, s4.1338

What we did with this particular one is we did kind of some algebraic cleverness.1343

In breaking n/n+2 up using partial fractions.1348

That gave us an expression that when we wrote out the partial sums,1355

They all cancelled in the middle and just left us with these beginning terms, and these ending terms.1360

This happens often and it is called a telescoping series.1366

It is called a telescoping series when you write out the sequence of partial sums and the middle terms all cancel,1381

Leaving you with just the terms at the beginning and the end.1391

Once you simplify it down to something in terms of the beginning and the end,1392

You can take the limit, whatever limit you get is, the limit of the sequence of partial sums.1399

By definition that is also the limit of the series.1405

We will try some more examples later.1410

This is educator.com.1413