For more information, please see full course syllabus of College Calculus: Level II

For more information, please see full course syllabus of College Calculus: Level II

### Series

**Main definitions and theorem:**

**Definitions:**

The notation stands for the series

*a*_{0}+*a*_{1}+*a*_{2}+ …The sequence of partial sums of the series is the sequence

The phrase the

__series__∑*a*_{n}__converges (to a limit__means that the*L*)__sequence of partial sums__{*s*_{n}}__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__ {*a*_{n}} 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__{*a*_{n}}, the__series__∑*a*_{n}, and the__sequence of partial sums__{*s*_{n}}. 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 ratioassuming the ratio has absolute value less than 1.

### Series

_{n = 1}

^{∞}7(2)

^{n − 1}converge or diverge?

_{n = 1}

^{∞}[(3

^{n})/3] converge or diverge?

- Simplify term
- ∑
_{n = 1}^{∞}[(3^{n})/3] = ∑_{n = 1}^{∞}3^{n − 1}

_{n = 1}

^{∞}18( [1/3] )

^{n − 1}converge or diverge?

_{n}= 7(2)

^{n − 1}

- Use the Sum of the first n terms equation
- S
_{n}= [(a(1 − r^{n}))/((1 − r))] - Define variables
- a = 7
- r = 2
- n = 9
- Apply equation
- S
_{n}= [(a(1 − r^{n}))/((1 − r))] - = [(7(1 − 2
^{9}))/((1 − 2))]

_{n}= 18( [1/3] )

^{n − 1}

- Use the Sum of the first n terms equation
- S
_{n}= [(a(1 − r^{n}))/((1 − r))] - Define variables
- a = 18
- r = [1/3]
- n = 3

_{n}= [(a(1 − r

^{n}))/((1 − r))] = [(18( 1 − ( [1/3] )

^{3}))/(( 1 − [1/3] ))] = 26

1 + [4/5] + [16/25] + [64/125] + ... = 1 + [4/5] + [(4

^{2})/(5

^{2})] + [(4

^{3})/(5

^{3})]a

_{n}= ( [4/5] )

^{n − 1}

- Use the Sum of the first n terms equation
- S
_{n}= [(a(1 − r^{n}))/((1 − r))] - Define variables
- a = 1
- r = [4/5]
- n = 5

_{n}= [(a(1 − r

^{n}))/((1 − r))] = [(( 1 − ( [4/5] )

^{5}))/(( 1 − [4/5] ))] = 3.36

- Use the Sum of series equation
- S = [a/(1 − r)]
- Define variables
- a = 1
- r = [4/5]

_{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])]

*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

### College Calculus 2 Online Course

I. Advanced Integration Techniques | ||
---|---|---|

Integration by Parts | 24:52 | |

Integration of Trigonometric Functions | 25:30 | |

Trigonometric Substitutions | 30:09 | |

Partial Fractions | 41:22 | |

Integration Tables | 20:00 | |

Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule | 22:36 | |

Simpson's Rule | 21:08 | |

Improper Integration | 44:18 | |

II. Applications of Integrals, part 2 | ||

Arclength | 23:20 | |

Surface Area of Revolution | 28:53 | |

Hydrostatic Pressure | 24:37 | |

Center of Mass | 25:39 | |

III. Parametric Functions | ||

Parametric Curves | 22:26 | |

Polar Coordinates | 30:59 | |

IV. Sequences and Series | ||

Sequences | 31:13 | |

Series | 31:46 | |

Integral Test | 23:26 | |

Comparison Test | 22:44 | |

Alternating Series | 25:26 | |

Ratio Test and Root Test | 33:27 | |

Power Series | 38:36 | |

V. Taylor and Maclaurin Series | ||

Taylor Series and Maclaurin Series | 30:18 | |

Taylor Polynomial Applications | 50:50 |

### 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, s _{1}, 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 s _{1} is just the s_{1} term, ln(2) - ln(1),*0077

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

*s _{3}, 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, s _{4}.*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 3 ^{n}/4^{3n+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 3 ^{2}/4^{3×2 + 2}.*0019

*So that is 4 ^{8}.*0028

*Now n=3 gives us 3 ^{3}, over 4^{3×3+2}, so that is 4^{11},*0031

*+ 3 ^{4}/4^{3n+2}, so 4^{14}.*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 × 3/4 ^{3}.*0064

*To get from the second to the third term, we multiply by 3/4 ^{3}.*0073

*So, each one of these terms, you get it by multiplying the previous term by 3/4 ^{3}.*0082

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

*Our common ratio is r = 3/4 ^{3}.*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 3/4 ^{3} < 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 3 ^{2}/4^{8}.*0170

*The common ratio is 3/4 ^{3}.*0176

*So, that is a little bit messy.*0186

*We can clean it up a little bit by multiplying top and bottom by 4 ^{3}.*0189

*That will give us 3 ^{2}/4^{5}.*0195

*On 4 ^{3} - 3 in the denominator.*0199

*That is 9/4 ^{5}/64-3.*0205

*So, that in turn becomes 9/61×4 ^{5}.*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.com*0000

*We are going to talk today about series*0003

*There are several bits of notations and definitions before we look at some examples*0005

*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 a _{0}, a_{0} + 1, a_{0} + 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 a _{0} + a_{1},the third is a_{0} + a_{1} + a_{2} and so on.*0047

*The s _{n} is a_{0} + a_{1} up to a_{n}*0054

*You call this the sequence of partial sums.*0060

*s _{0}, s_{1}, s_{2} up to s_{n}.*0064

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

*This is a sequence s _{n}.*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 sequence*0102

*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 × r ^{2} 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 value*0270

*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/n*0513

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

*a _{n} = n-1/n.*0520

*We rewrite that as 1-1/n.*0525

*The limit of the sequence a _{n} 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 a _{n} 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

*a _{n} = 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

*s _{1} is just the first partial sum, that is just 1.*0676

*s _{2} is 1 + 1/2.*0679

*s _{3} is 1 + 1/2 + 1/3.*0684

*s _{4} 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 s _{3}, I will not look at that.*0726

*s _{4}, 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 s _{n}.*0756

*s _{n} 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 s _{16}.*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 s _{n} 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-n ^{p} where p=1.*0952

*It is really 1/n ^{1}.*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

*s _{1}, 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

*s _{2} is 1/2 × 1 - 1/3, + the n = 2 term, is 1/2 - 1/4.*1127

*s _{3} 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, s _{4}.*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 s _{n}.*1216

*It is going to be 1/2, well after we cancel everything, the only terms left are this 1*1218

*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 s _{1}, s_{2}, s_{3}, s_{4}.*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

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!