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

## Discussion

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## Table of Contents

## Transcription

## Related Books

### Comparison Test

**Main theorems:**

Suppose and are series with __positive terms__. ( is given to you, and you create yourself.)

**Comparison Test:**

If

*a*_{n}<*b*_{n}for all*n*and converges, then converges.If

*a*_{n}>*b*_{n}for all*n*and diverges, then diverges.

**Limit Comparison Test:** Suppose lim = some finite, positive (not 0) number.

Then if converges then converges, and

if diverges then diverges.

**Hints and tips:**

Like the Integral Test, the Comparison Tests only work for series with

__positive__terms. However, after you learn about absolute convergence later, you may be able to use them for series with some negative terms by taking their absolute value and seeing if they are absolutely convergent.The idea with these tests is that you are given a series . You come up with the series yourself. It should be similar in form to , but simpler to analyze.

Often you will take to be a p-series or geometric series. Remember the differences between these two and the conditions under which each one converges or diverges.

The Comparison Tests are two-way tests − you can use them to conclude that a series converges or that it diverges. However, it’s very important that the inequalities go the right way. If you have a series that is bigger than a known convergent series, the Comparison Test tells you nothing.

And if you have a series that is smaller than a known divergent series, the Comparison Test tells you nothing.

Because the inequalities must go the right way, it’s often useful to get some intuition about whether a series converges or diverges before setting up a comparison. It’s useful to remember the ranking of functions:

When you have a complicated rational expression, focus on the biggest term in the numerator and the biggest term in the denominator.

### Comparison Test

_{}

^{}[1/(2 + n

^{4})] converge?

- Determine an appropriate u
_{n}for the Comparison Test - Consider u
_{n}= [1/(n^{4})] - Consider n = 2, then a
_{n}≤ u_{n}

_{}

^{}[1/(n

^{4})] p = 4 > 1, thus the series u

_{n}converges.

Since u

_{n}converges, then a

_{n}converges

_{}

^{}[1/(n

^{2}− 50)] converge?

- Determine an appropriate u
_{n}for the Comparison Test - Consider u
_{n}= [1/(n^{2})] - Consider n = 3, then a
_{n}≤ u_{n}

_{}

^{}[1/(n

^{2})] p = 2 > 1, thus the series u

_{n}converges

Since u

_{n}converges, then a

_{n}converges

_{}

^{}[1/(√{n + 6})] converge?

- Determine an appropriate u
_{n}for the Comparison Test - Consider u
_{n}= [1/(n^{1/2})] - Consider n = 3, then a
_{n}u_{n} - Determine ∑
_{}^{}u_{n}diverges with P - series properties - ∑
_{}^{}[1/(n^{1/2})] - p = [1/2] ≤ 1, thus the series u
_{n}diverges

_{n}diverges, then a

_{n}diverges

- Represent the series as a summation
- 1 + [1/2] + [1/(√7 )] + ... + [1/(√{3n − 2} )] + ... = ∑
_{}^{}[1/(√{3n − 2} )] - Determine an appropriate u
_{n}for the Comparison Test - Consider u
_{n}= [1/(√{3n} )] = [1/(√3 )]( [1/(√n )] ) - Consider n = 2, then a
_{n}u_{n}

_{}

^{}[1/(√{3n} )] = ∑

_{}

^{}[1/(√3 )]( [1/(√n )] ) = [1/(√3 )]∑

_{}

^{}[1/(√n )] p = [1/2] ≤ 1, thus the series u

_{n}diverges

Since u

_{n}diverges, then a

_{n}diverges

_{}

^{}u

_{n}diverges and a

_{n}≤ u

_{n}, does ∑

_{}

^{}a

_{n}diverge?

No, because the condition requires a

_{n}u

_{n}

_{}

^{}[(n + 1)/(n

^{2}+ 2)] converge?

- Determine an appropriate u
_{n}for the Comparison Test - Consider u
_{n}= [1/(n^{})] - Consider n = 3, then a
_{n}u_{n}

_{}

^{}[1/(n

^{})] Since it's a Harmonic Series, the series u

_{n}diverges

Since u

_{n}diverges, then a

_{n}diverges

_{}

^{}[1/(√{n

^{3}+ 6} )] converge?

- Determine an appropriate u
_{n}for the Comparison Test - Consider u
_{n}= [1/(√{n^{3}} )] - Consider n = 3, then a
_{n}≤ u_{n} - Determine ∑
_{}^{}u_{n}converges with P − series properties - ∑
_{}^{}[1/(√{n^{3}} )] = ∑_{}^{}[1/(n^{3/2})] - p = [3/2] > 1, thus the series u
_{n}converges

_{n}converges, then a

_{n}converges

_{}

^{}u

_{n}converges and a

_{n}u

_{n}, does ∑

_{}

^{}a

_{n}converge?

No, because the condition requires a

_{n}≤ u

_{n}

_{}

^{}[(n

^{3})/((n + 1)

^{2})] converge?

- Determine an appropriate u
_{n}for the Comparison Test - Consider u
_{n}= n - Consider n = 3, then a
_{n}≤ u_{n}

_{}

^{}n = ∞ Thus the series u

_{n}diverges

Since u

_{n}diverges, then a

_{n}diverges

_{}

^{}[1/(5n

^{3}+ n

^{2}+ 1)] converge?

- Determine an appropriate u
_{n}for the Comparison Test - Consider u
_{n}= [1/(n^{3})] - Consider n = 2, then a
_{n}≤ u_{n} - Determine ∑
_{}^{}u_{n}converges with P - series properties - ∑
_{}^{}[1/(n^{3})] - p = 3 > 1, thus the series u
_{n}converges

_{n}converges, then a

_{n}converges

- Use b
_{n}= [1/n] for Limit Comparison Test

_{n}diverges

- Use b
_{n}= [1/n] for Limit Comparison Test

_{n}diverges

- Use b
_{n}= [1/(√n )] for Limit Comparison Test

_{n}converges

^{2}+ 2n + 1)] converge?

- Use b
_{n}= [1/(n^{2})] for Limit Comparison Test - lim
_{n → ∞}[([1/(n^{2}+ 2n + 1)] )/([1/(n^{2})])] = lim_{n → ∞}[(n^{2})/(n^{2}+ 2n + 1)] - = [(n
^{2})/(n^{2})] - = 1

_{n}converges

_{n}= n

^{3}+ 3n if b

_{n}= n

^{2}?

No it is not appropriate because lim

_{n → ∞}[(a

_{n})/(b

_{n})] is infinite

^{3}− 1)] converge?

- Use b
_{n}= [1/(n^{2})] for Limit Comparison Test - lim
_{n → ∞}[([(n − 2)/(n^{3}− 1)])/([1/(n^{2})])] = lim_{n → ∞}[(n^{2}( n − 2 ))/(n^{3}− 1)]

^{3})/(n

^{3})] = 1 Thus the series converges because b

_{n}converges

^{2})] diverge?

- Use b
_{n}= [1/(n^{3/2})] for Limit Comparison Test - lim
_{n → ∞}[([(√n )/((2n − 1)^{2})])/([1/(n^{3/2})])] = lim_{n → ∞}[(√n (n^{3/2}))/((2n − 1)^{2})]

^{2})/(4n

^{2})] = [1/4] Thus the series converges because b

_{n}converges

^{2})/(9n

^{3}+ 8n

^{2}+ 2n + 3)] diverge?

- Use b
_{n}= [1/n] for Limit Comparison Test - lim
_{n → ∞}[([(7n^{2})/(9n^{3}+ 8n^{2}+ 2n + 3)])/([1/n])] = lim_{n → ∞}[(7n^{3})/(9n^{3}+ 8n^{2}+ 2n + 3)] - = [(7n
^{3})/(9n^{3})] - = [7/9]

_{n}diverges

^{2}+ n)/(5n

^{3}− 12n

^{2}− n + 17)] diverge?

- Use b
_{n}= [1/n] for Limit Comparison Test

^{3})/(5n

^{3})] = [3/5] Thus the series diverges because b

_{n}diverges

- Analyze the series
- 3 + [3/4] + [1/3] + [3/16] + ... = ∑[3/(n
^{2})] - Use b
_{n}= [1/(n^{2})] for Limit Comparison Test - lim
_{n → ∞}[([3/(n^{2})])/([1/(n^{2})])] = lim_{n → ∞}[(3n^{2})/(n^{2})] - = 3

_{n}converges

*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

### Comparison Test

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 Tests 0:01
- Comparison Test
- Limit Comparison Test
- Lecture Example 1 1:44
- Lecture Example 2 3:52
- Lecture Example 3 6:01
- Lecture Example 4 10:04
- Additional Example 5
- Additional Example 6

### 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: 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 e ^{n}/n!.*0005

*That one is a little tricky.*0011

*I want to write that one as e ^{n}/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 = e ^{n}/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 3 ^{2}.*0064

*3 ^{2} × e^{n}/1 × 2, now I can say that is 3^{n}.*0071

*I just multiplied in 3 ^{2} × 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, 3 ^{2} is 9/2 × e^{n}/3^{n}.*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/3^{n}.*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 test*0175

*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 n ^{4}+3n-8/3n^{7}+6n^{5}.*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, n ^{4}, in the numerator, and n^{7} in the denominator.*0024

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

*It looks like the sum of n ^{4}/n^{7}.*0038

*That in turn would simplify down to 1/n ^{3}.*0041

*OK, I am going to call my 1/n ^{3}, 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 n^{4}+3n-8/3n^{7}+6n^{5}.*0069

*b _{n} is 1/n^{3},*0081

*But we can flip that up into the numerator so we can write that as n ^{3}/1.*0084

*That turns into n ^{7}+3n^{4}-8n^{3}/3n^{7}+6n^{5}.*0094

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

*Which is an n ^{7} in the top and the bottom and you divide by it.*0113

*So, you get 1+3/n ^{3}-8/n^{4}/3+6/n^{2}.*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/n ^{3}, 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.com*0259

*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

*3 ^{n}/2^{n} - 1,*0233

*That looks like the sum of 3 ^{n}/2^{n}.*0240

*So a _{n} is the given series, for our b_{n}, we will use the sum of 3^{n}/2^{n}.*0249

*Then we will compare those to each other.*0255

*So, 3 ^{n}/2^{n}-1 vs. 3^{n}/2^{n}.*0260

*Well 2 ^{n}-1 is a smaller denominator which means that 3^{n}/2^{n}-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/2 ^{n} is just 3/2^{n}, 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(2n ^{2}) + 17n/n.*0444

*If we look at the top and bottom there, the biggest terms we have in the top, we have an n ^{2} 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 n ^{2}.*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/-n ^{2}.*0707

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

*× derivative of 1/n by the chain rule which is -1/n ^{2}.*0718

*Those -1/n ^{2}'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

1 answer

Last reply by: Dr. William Murray

Thu Aug 4, 2016 6:02 PM

Post by Peter Ke on July 30, 2016

For example 3, why can't we go straight to testing if the exponent is bigger than or less than 1 instead of wasting time doing the limit comparison test?

1 answer

Last reply by: Dr. William Murray

Thu Apr 24, 2014 6:18 PM

Post by Brandyn Albrecht on April 22, 2014

Quick question. taking an/bn is just to figure out if the series are similar or whatever in terms of divergence, but for example 1 just to double check I took (1/n+1)/1/n and got n/(n+1) which converges to 1, which is a positive real number, which means an and bn should act similarly right? But they don't because an is smaller than bn, so does the an/bn approach only work for more confusing questions like example 3, and not for all questions in general?

3 answers

Last reply by: Dr. William Murray

Tue Dec 17, 2013 9:09 PM

Post by Xenia Jeanty on November 30, 2013

lecture example 3 why is sqrt n/n same as 1/sqrt n? Can you please explain how did we get there?

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Last reply by: Dr. William Murray

Tue Aug 13, 2013 5:24 PM

Post by charles daniel on July 24, 2013

lecture example 3 in the denominator n/n^2 should be equal to zero ..right?..why is it equal to one

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Last reply by: Dr. William Murray

Tue Aug 13, 2013 5:21 PM

Post by charles daniel on July 24, 2013

lecture example 3 .. how is the bn of root(2n+17)/n become root n/n

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Last reply by: Dr. William Murray

Tue Apr 16, 2013 8:27 PM

Post by Alena Schwartsman on April 14, 2013

Thank you Dr. Murray, concise and to the point. I like how you repeat the test's requirements with every example; easier to memorize the tests this way. Overall, I love how you explain things! Thank you again!

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Last reply by: Dr. William Murray

Wed Apr 3, 2013 6:43 PM

Post by Rohail Tariq on April 1, 2013

cool

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Last reply by: Dr. William Murray

Wed Apr 3, 2013 6:42 PM

Post by Ahmad Alshammari on September 13, 2011

why you put .. 3^(n-2) !!

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Last reply by: Dr. William Murray

Wed Apr 3, 2013 6:34 PM

Post by Srinivasa Rao RAVELLA on April 14, 2011

how would you should tan(1/x) is convergent with the limit comparison test

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Last reply by: Dr. William Murray

Wed Apr 3, 2013 6:26 PM

Post by Romin Abdolahzadi on April 4, 2011

Regarding the limit comparison test: If the limit is equal to 0, then if Bn converges then An also converges!