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

### Ratio Test and Root Test

**Main definitions and theorems:**

**Definitions:** Let ∑*a _{n}* be a series.

∑

*a*_{n}__is Absolutely Convergent__means that ∑*a*converges and ∑|_{n}*a*| converges._{n}∑

*a*_{n}__is Conditionally Convergent__means that ∑*a*converges but ∑|_{n}*a*| diverges._{n}∑

*a*_{n}__is Divergent__means that ∑*a*diverges and ∑|_{n}*a*| diverges._{n}

**Theorem:** If ∑|*a _{n}*|
converges, then ∑

*a*converges.

_{n}**Ratio Test:** Suppose ∑*a _{n}*
is a series. Calculate lim
and call it

*L*.

If

*L*< 1, then the series__converges absolutely__.If

*L*> 1, then the series__diverges__.If

*L*= 1, then the Ratio Test doesn’t tell you. The series could converge absolutely, converge conditionally, or diverge.

**Root Test:** Suppose ∑*a _{n}*
is a series. Calculate lim
and call it

*L*.

If

*L*< 1, then the series__converges absolutely__.If

*L*> 1, then the series__diverges__.If

*L*= 1, then the Root Test doesn’t tell you. The series could converge absolutely, converge conditionally, or diverge.

**Hints and tips:**

Think of an Absolutely Convergent series as being “super-convergent”. It is so strongly convergent that it still converges even if we make everything positive.

Think of a Conditionally Convergent series as being “just barely convergent”. It only converges because some of the terms are negative.

Here are some flags that might suggest the Ratio Test:

(−1)

^{n}, since many other tests require positive terms.*n*!, (2*n*)!, since factorials cancel nicely in ratios. (Remember that (*n*+ 1)! = (*n*+ 1)*n*! and [2(*n*+ 1)]! = (2*n*)!(2*n*+ 1)(2*n*+ 2).)constant

^{n}(including*e*^{n})Any kind of power series/Taylor Series (coming later)

Here are some flags that might suggest that you not use the Ratio Test:

*n*,*n*² (or any polynomial), √*n*, ln*n*. All of these have lim = 1, so the Ratio Test fails. But if you have some of this clutter mixed with some good stuff, Ratio is great for clearing the clutter away and leaving you with something good.*n*, or more generally (a function of^{n}*n*)^{(another function of n)}(more appropriate for Root Test)Remember that .

Here are some flags that might suggest the Root Test:

(−1)

^{n}, since many other tests require positive terms.(a function of

*n*)^{(another function of n) }, e.g.*n*, (^{n}*n*− 2)²^{n}, etc. Not (constant)^{n}(use geometric series or Ratio Test) or*n*^{constant}(use p-series or LCT).Here are some flags that might suggest that you

__not__use the Root Test:*n*!, (2*n*)!, etc. (Use Ratio Test.)*n*² ,*n*, √*n*. All of these have lim |*a*|_{n}^{1/n}= 1. But if you have some of this mixed with some stuff that looks good for Root Test, then you can use Root.Remember that you can write . Then you can often sort out the exponent using L’Hôpital.

It is probably worth memorizing that lim =

*e*, or even that

lim .Also remember that

*e*≈ 2.7, so*e*^{k}> 1 if*k*is positive and*e*^{k}< 1 if*k*is negative.

(Note that the Ratio Test is based on
comparison with geometric series. So things that look geometric,
e.g. constant^{n} , make it work. p-series do
not.)

### Ratio Test and Root Test

^{n}converges or diverges

- Apply the Ratio Test, find a
_{n + 1} - a
_{n + 1}= 3^{n + 1}

_{n → ∞}[(3

^{n + 1})/(3

^{n})] = 3L > 1, thus ∑3

^{n}diverges

^{n})] converges or diverges

- Apply the Ratio Test, find a
_{n + 1} - a
_{n + 1}= [1/(2^{n + 1})] - Find and analyze L

_{n → ∞}[(2

^{n})/(2

^{n + 1})] = [1/2]L < 1, thus ∑[1/(2

^{n})] converges

^{n})] converges or diverges

- Apply the Ratio Test, find a
_{n + 1} - a
_{n + 1}= [(n + 1)/(3^{n + 1})] - Find and analyze L
- L = lim
_{n → ∞}[([(n + 1)/(3^{n + 1})])/([n/(3^{n})])] - = lim
_{n → ∞}[(n + 1)/(3^{n}(3))]( [(3^{n})/n] ) - = lim
_{n → ∞}[(n + 1)/3n] - = [n/3n]
- = [1/3]

^{n})] converges

^{n})/(n

^{2})] converges or diverges

- Apply the Ratio Test, find a
_{n + 1} - a
_{n + 1}= [(5^{n + 1})/((n + 1)^{2})] - = [(5
^{n + 1})/(n^{2}+ 2n + 1)] - Find and analyze L
- L = lim
_{n → ∞}[([(5^{n + 1})/(n^{2}+ 2n + 1)])/([(5^{n})/(n^{2})])] - = lim
_{n → ∞}[(5^{n + 1})/(n^{2}+ 2n + 1)]( [(n^{2})/(5^{n})] )

^{2})/(n

^{2})] = 5L > 1, thus the series diverges

- Apply the Ratio Test, find a
_{n + 1} - a
_{n + 1}= [(n + 2)/n] - Find and analyze L
- L = lim
_{n → ∞}[([(n + 2)/n])/([(n + 1)/(n − 1)])] - = lim
_{n → ∞}[(n + 2)/n]( [(n − 1)/(n + 1)] )

^{2})/(n

^{2})] = 1L = 1, thus another test should be applied.

- Apply the Ratio Test, find a
_{n + 1} - a
_{n + 1}= [2/((n + 2)!)] - Find and analyze L
- L = lim
_{n → ∞}[([2/((n + 2)!)])/([2/((n + 1)!)])] - = lim
_{n → ∞}[2/((n + 2)!)]( [((n + 1)!)/2] ) - = lim
_{n → ∞}[1/(n + 2)] - = 0

^{n})/n!] converges or diverges

- Apply the Ratio Test, find a
_{n + 1} - a
_{n + 1}= [(2^{n + 1})/((n + 1)!)] - Find and analyze L
- L = lim
_{n → ∞}[([(2^{n + 1})/((n + 1)!)])/([(2^{n})/n!])]

_{n → ∞}[2/(n + 1)] = 0L < 1, thus the series converges

- Apply the Ratio Test, find a
_{n + 1} - a
_{n + 1}= [((n + 1)!)/((n − 9)!)] - Find and analyze L
- L = lim
_{n → ∞}[([((n + 1)!)/((n − 9)!)])/([n!/((n − 10)!)])] - = lim
_{n → ∞}[((n + 1)!)/((n − 9)!)]( [((n − 10)!)/n!] ) - = lim
_{n → ∞}( n + 1 )(n − 10) - = ∞

^{n})/(6

^{n})] converges or diverges

- Apply the Ratio Test, find a
_{n + 1} - a
_{n + 1}= [(4^{n + 1})/(6^{n + 1})] - Find and analyze L
- L = lim
_{n → ∞}[([(4^{n + 1})/(6^{n + 1})])/([(4^{n})/(6^{n})])]

_{n → ∞}[4/6] = [4/6]L < 1, thus the series converges

^{n + 1})/(3

^{n − 1})] converges or diverges

- Apply the Ratio Test, find a
_{n + 1} - a
_{n + 1}= [(2^{n + 2})/(3^{n})] - Find and analyze L
- L = lim
_{n → ∞}[([(2^{n + 2})/(3^{n})])/([(2^{n + 1})/(3^{n − 1})])] - = lim
_{n → ∞}[(2^{n + 2})/(3^{n})]( [(3^{n − 1})/(2^{n + 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

### Ratio Test and Root 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
- Theorems and Definitions 0:06
- Two Common Questions
- Absolutely Convergent
- Conditionally Convergent
- Divergent
- Missing Case
- Ratio Test 3:07
- Root Test 4:45
- Lecture Example 1 5:46
- Lecture Example 2 9:23
- Lecture Example 3 13:13
- 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: Ratio Test and Root Test

*Hi we are here trying more examples of the ratio test and the root test.*0000

*We have a kind of complicated one, -100 ^{n} n^{4}/n! × ln(n).*0004

*That one looks a little nasty but you will see that the ratio test handles it pretty well.*0014

*So, ratio test says you look at a _{n}+1/a_{n}.*0022

*I am going to work out this whole expression, but substituting an n+1 for n.*0028

*We get -100 ^{n+1} (n+1)^{4}/n+1! × ln(n+1).*0036

*All of that divided by the same thing just with n, so -100 ^{n} n^{4}/n! × ln(n).*0060

*Again, I am going to flip the denominator and bring it up and multiply it into the numerator,*0070

*I am going to go ahead and pair the factors off with factors that look like them.*0078

*We get 100 ^{n+1}/100^{n}.*0084

*By the way, I am getting rid of the negative signs because we are taking absolute values there.*0089

*n+1/4/n ^{4}.*0096

*Let us see, this n! is going to flip up into the numerator,*0103

*So we have n!/n+1!.*0109

*Then the ln(n) is going to flip up into the numerator.*0112

*ln(n) × ln(n+1).*0117

*Then let us look at each one of those factors separately.*0122

*100 ^{n+1}/100^{n} is 100.*0126

*Now (n+1) ^{4} is going to be some big complicated polynomial.*0130

*The key thing here is that n ^{4} plus some smaller terms.*0136

*We are dividing it by n ^{4},*0145

*So both in top and bottom there, the biggest term is n ^{4}.*0148

*We can divide top and bottom by n ^{4},*0154

*So, that will make the n ^{4} terms go to 1 when you take the limit.*0161

*All of the smaller terms go to 0.*0168

*This just goes to 1 as n goes to infinity.*0171

*Remember, that was the pattern we saw before with polynomials.*0177

*When you take the ratio of n+1 and n in a polynomial, you always get 1.*0180

*That is very useful for the ratio test.*0186

*n+1!, remember we can write that as n!.*0189

*× n+1, so the n! cancels with the n! in the numerator.*0195

*We get 1/n+1.*0201

*Then ln(n)/ln(n+1).*0204

*For that we could say that that goes to infinity/infinity.*0209

*We could use l'hopital's rule.*0215

*L'hopital's rule says you take the derivative of the top and bottom,*0219

*That is 1/n divided 1/n+1.*0220

*So that is n+1/n.*0226

*The limit of that as n goes to infinity is just 1.*0231

*You put all of these things together and you get, 100/n+1.*0240

*Now, you can take the limit as n goes to infinity.*0249

*Now 100 is constant and n is getting bigger and bigger, so this limit is 0.*0255

*We can say l=0, so the series converges absolutely.*0259

*By the ratio test.*0280

*The key thing about the 0 there is that it is less than 1.*0288

*That was the cut off to check for the ratio test.*0295

*There is a lot of things we can learn from this example.*0300

*If you look at the different terms here, which ones work well for the ratio test?*0305

*Well, the -100, if we follow the -100,*0312

*If we kind of trace those terms through the ratio test,*0317

*That gave us a 100, which was very very good for the ratio test.*0321

*I like using that for the ratio test.*0328

*Good in the sense that it did contribute to the answer.*0330

*n ^{4}, if you look at n^{4} and follow that through,*0335

*That just gave us 1 at the end.*0340

*Remember, in the ratio test, if you get 1 as an answer, it does not tell you whether or not the series converges.*0343

*So n ^{4} was not useful for the ratio test.*0350

*n!, we have got n+1! and n!, that cancelled away nicely, we got 1/n+1,*0355

*Which definitely had an influence on the answer,*0366

*So the n! was good for the ratio test.*0370

*ln(n), if you trace those through, ln(n), just ended up giving us 1.*0374

*Remember, if you get 1 as the answer to the ratio test, it gives you no information.*0383

*If you have just things like polynomials like this n ^{4}, or natural logs,*0388

*Your answer for the ratio test is going to be 1 which gives you no information and it means you did a lot of work for nothing.*0396

*On the other hand, if you have geometric terms, like -100 ^{n},*0405

*Or factorial terms, like n!,*0410

*Then the ratio test is very useful in giving an answer.*0415

*Let me summarize that there.*0418

*If you have geometric terms, like 100 ^{n}, or factorial terms,*0421

*You definitely want to use the ratio test.*0434

*Because it will give you a useful answer.*0439

*On the other hand, if you have polynomial terms,*0444

*Or natural log terms, remember polynomials like n ^{4},*0450

*When you run those through the ratio test, what you get for your answer is 1.*0456

*The ratio test gives you no information if your answer comes out to be 1.*0463

*If you see those turns, do not use the ratio test.*0470

*You will be wasting your time and you will not get an answer.*0473

*Do not use the ratio test on those terms.*0480

*Unless you also have some of the good terms, the geometric and factorial stuff.*0483

*Good factors.*0502

*To summarize that, the ratio test works very well on geometric and factorial stuff.*0506

*It works badly on polynomial and natural log stuff.*0512

*However, if you have a mix of both,*0516

*You do want to use the ratio test.*0520

*Because the ratio test will sort of sweep away the polynomial and the natural log stuff.*0523

*It will leave you with the good stuff, the geometric stuff and the factorial stuff that will actually give you a meaningful answer.*0530

*That is kind of when to use the ratio test and when not to use the ratio test.*0541

*How you can tell ahead of time whether the ratio test is going to be a useful one,*0546

*Or whether it is going to be a big waste of time.*0550

*We are going to try one more example here.*0000

*The sum of 1-2/n ^{n2}.*0004

*What I notice right away about this one is that I have some stuff involving n, the 1-2/n,*0010

*Raised to some stuff involving n in the exponent.*0016

*I have n both in the base and the exponent.*0022

*That is not something I want to use the ratio test.*0027

*Do not use the ratio test for something like that.*0028

*Instead, you want to use the root test for something like that.*0032

*When you have something involving n raised to something involving n in the exponent,*0039

*You want to use the root test on that.*0044

*Let us try this out using the root test.*0047

*Remember that says you look at the absolute value of a _{n}^{1/n},*0050

*That is the nth root of the terms.*0055

*This is 1-2/n raised to the n ^{2}, all raised to the 1/n power.*0059

*I am going to get rid of the absolute values right away because 1-2/n,*0074

*After n gets big enough, that is going to be a positive number.*0080

*This is 1-2/n.*0085

*Now n ^{2} raised to the 1/n,*0090

*Remember you multiply those exponents.*0092

*That is n ^{2}/n, that is just n.*0097

*Now this is a tricky expression, 1-2/n ^{n}.*0103

*We have seen an example like this before in another lecture. *0108

*The trick here is when you have something complicated raised to a complicated exponent,*0112

*You try to separate them out.*0117

*You write this as e ^{ln(ab)}, the point of doing that is you can pull the exponent b out of the natural log.*0121

*You can write it as e ^{b}, ln(a).*0132

*Here the a is the 1-2/n and the b is n.*0138

*We can write this as e ^{n} × ln(1-2/n).*0142

*Now we are just going to focus on the exponent.*0155

*Because that will take a little work to sort out.*0158

*We will come back later and incorporate the e term.*0162

*Let us just look at that exponent.*0165

*n × ln(1-2/n).*0166

*n × ln(1-2/n), well n goes to infinity.*0175

*ln(1-2/n), well 2/n goes to 0.*0180

*That is ln(1) which is 0.*0186

*We have an infinity × 0 situation.*0189

*You cannot do anything with infinity × 0, it could come out to be anything.*0193

*What I mean is you cannot do anything directly.*0197

*What you can do is write it as ln(1-2/n)/1/n.*0200

*That is going to, the numerator is still 0, 1/n is now 0, that is going to 0 over 0.*0214

*That is a situation where you can use l'Hopital's rule.*0223

*We are going to invoke l'Hopital's rule here.*0228

*Remember l'Hopital's rule, you take the derivative of the top and bottom separately.*0233

*It is not like the quotient rule from Calculus 1.*0241

*It is kind of different.*0243

*You take the derivative of the top and bottom separately.*0244

*I will take the derivative of the bottom first because it is easier.*0248

*Derivative of 1/n is -1/n ^{2},*0250

*Now the derivative of the top, natural log of something,*0256

*Is 1/that something, 1-2/n.*0261

*× derivative of the inside part.*0263

*That is the derivative of 1-2/n is - 2/n ^{2}, and the negative of that turns that into a positive.*0267

*The derivative of 2/n is -2/n ^{2}, derivative of -2/n is positive 2/n^{2}.*0283

*Let us clean this up a little.*0294

*The 1/n ^{2} cancels with that 1/n^{2} and we are left with negative from the denominator 2/1-2/n.*0296

*Now, as n goes to infinity, the 2/n goes to 0.*0307

*This whole thing goes to -2.*0317

*Remember, this was the exponent.*0320

*If we incorporate back in this e that we had before,*0324

*The true limit here, a _{n} to the 1/n goes to e^{-2}.*0329

*So, our limit l that we checked for the root test is e ^{-2}.*0340

*That is the same as 1/e ^{2}.*0347

*We have to figure out approximately how big that is.*0350

*e is approximately 2.7.*0354

*1/e ^{2}, I do not know exactly what it is, but I know that it is definitely less than 1.*0357

*That is the cut off for the root test, and so,*0365

*If l < 1, that tells us that the series aN is absolutely convergent by the root test.*0370

*Let us recap what we did here.*0402

*The first thing we did when we looked at this series was to say,*0407

*I see an n in the base and an n in the exponent.*0410

*That is something complicated, not something that the ratio test is going to work well on.*0414

*That tells me right away, try for the root test.*0421

*I set up aN to the 1/n.*0425

*That simplifies down a little bit and we get to this step right here.*0428

*When we look at that step, again looks like something a little complicated.*0433

*I can sort it out using this e ^{ln} rule.*0439

*You sort it out using this e ^{ln} rule,*0442

*And if you look at the exponent, you get infinity × 0.*0447

*You can rewrite that as 0/0.*0451

*Which allows you to invoke l'Hopital's rule.*0454

*You work l'Hopital's rule through and you get -2.*0459

*But that -2 is actually just the exponent because you still have this e from earlier.*0462

*So you get e ^{-2} as our final limit.*0470

*The important thing about that was that it was less than 1,*0472

*And, the root test says that if your limit is less than 1, then the whole series is absolutely convergent.*0477

*This is educator.com, thanks for watching.*0486

*Hi, this is educator.com and we are here to learn about the ratio test and the root test.*0000

*Now, there are some definitions we need before we get started.*0007

*From now on, before you are given a series,*0010

*You are going to be asked two different questions packed into one.*0014

*The first is whether the original series converges.*0017

*The second question is if you take the absolute value of each term of the series, does it still converge?*0021

*The point there is that you are making all of the terms positive,*0031

*That potentially makes the series bigger and might make it diverge.*0033

*There are three classifications that we are going to be using for series from now on.*0039

*We are going to say it is absolutely convergent if the original series converges and the absolute series converges.*0047

*The words absolutely convergent are supposed to make you think of something that is very strongly convergent.*0054

*In the sense that, I say super-convergent, even if you make all of the terms positive,*0064

*Even if you make all of the terms as big as you can, that series still converges.*0070

*That is absolute convergence.*0075

*Conditional convergence means that the series converges,*0078

*But the absolute series diverges.*0084

*The way you want to think about that is the series just barely converges.*0090

*It does converge, but if you make all of the terms positive,*0092

*Then it gets so big that it diverges.*0096

*It only just barely converged because some of the original terms were negative.*0100

*It had to have some of the negative terms to make it converge.*0105

*If you make all of the terms positive, it diverged.*0108

*Then finally, we will talk about divergent series,*0110

*Means that the series diverges and the absolute series also diverges.*0115

*Now if you think about this, there is one case missing from those 3 categories.*0122

*The case that is missing is what if a _{n} diverges,*0127

*But the absolute series converges.*0137

*Now, what about that possibility?*0145

*The answer is that that possibility cannot happen.*0148

*Essentially, you want to think about it as the absolute series is always at least as big as the original series.*0150

*If the absolute series converges, then the original series must converge as well.*0158

*That is what this theorem is saying.*0164

*If the absolute series converges, then the original series must converge as well.*0166

*You cannot have the situation where the original series diverges but the absolute series converges.*0172

*That never happens.*0178

*There are really only three possible answers for all the series that we are going to study today.*0180

*We have a couple tests to help you determine those.*0185

*You will be given a series,*0190

*And, one nice thing here is that you do not have to check anything about whether the series has positive terms or anything like that.*0191

*This is a very strong test.*0200

*It works for any kind of series.*0203

*For any kind of series where you can calculate this limit.*0206

*What you do is you look at the n+1 term, and you look at the n term,*0208

*You divide them together, you take the absolute value, and then you take the limit of that.*0213

*Whatever you get, you call it l.*0218

*Now if l is less than 1, then the series converges absolutely.*0221

*If l is bigger than 1, the series diverges.*0226

*The natural inclination at this point is to say that if l=1 then the series converges conditionally, because that is the remaining case.*0230

*That is not true.*0238

*If l=1, then it kind of breaks down.*0240

*The ratio test gives you no answer, and there are examples where the series converges absolutely.*0244

*There are other examples where the series converges conditionally.*0251

*There are others where it diverges.*0254

*Essentially if l=1, then you have wasted your time using the ratio test,*0256

*Because it does not give you an answer, you have to find something else.*0262

*That is the downside of the ratio test, if you get l=1.*0266

*However, if you get l = bigger than 1 or less than 1, that includes infinity and 0, *0268

*Those are OK to use here, then you do get a good answer.*0272

*You can say that the series converges absolutely, or diverges. That is the ratio test.*0278

*We will do some examples in a second but I want to introduce you also to the root test.*0284

*Which looks a lot like the ratio test but instead of calculating a ratio, you calculate a root.*0287

*You look at absolute value of the terms, and then you take the nth root.*0297

*Remember, that is the same as taking a _{n} to the 1/n power.*0302

*You take the limit again and you call that number l.*0306

*Then you have the exact same conclusions as the ratio test.*0310

*If l < 1, it converges absolutely, if l > 1, it diverges, and if l = 1, then you might hope to say it converges conditionally,*0314

*But that would be very unjustified.*0326

*You would have to look at some other tests,*0330

*Because just based on the root test, we do not know yet whether it converges conditionally, converges absolutely, or diverges.*0332

*Let us try that out with some examples.*0343

*You will see how it works.*0345

*First example I have is -e ^{n}/n!.*0347

*So we are going to go for the ratio test.*0351

*We are going to look at, remember a _{n} +1/a_{n} , and the absolute value of that.*0353

*aN+1 means you look at each term of the series and you plug in n+1 wherever you see n.*0360

*So that is -e ^{n+1}/n+1!.*0369

*a _{n} just means the original term, -e^{n}/n!.*0376

*Take absolute value of that.*0382

*What I am going to do is flip these two fractions.*0385

*Flip the denominator so we can compare it with a numerator.*0388

*I will write that as e ^{n+1}/e^{n} × n!/n+1!.*0393

*There are a couple things to explain here.*0403

*Notice with the n! and the e ^{n},*0406

*I got that just by flipping the denominator up and rearranging them to match their respective factors in the numerator.*0410

*The other thing that I noticed was you took out the negative signs and that is because of the absolute values.*0419

*The absolute values make everything positive so I can just throw away the negative signs there.*0425

*Now, let us work with this a bit.*0428

*e ^{n+1}/e^{n} is just e.*0431

*Now n!/n+1!, remember n+1 factorial.*0435

*That means you multiply all the numbers together, up to n × n+1.*0443

*You can write that as n! × n+1.*0452

*We will write that as n! × n+1, and now the n! cancel.*0458

*So you get, e/n+1.*0464

*Now, remember the ratio test says you take the limit of this as n goes to infinity.*0468

*We will take the limit as n goes to infinity.*0475

*e is just a number, it is 2.7, well 2.7 and on and on and on.*0480

*n+1 is getting bigger and bigger and bigger,*0487

*So we are dividing a constant by a bigger and bigger number.*0491

*That limit is 0. Our l=0.*0493

*Now the important thing in the ratio test is you check whether that limit is < or > 1.*0499

*0 is certainly less than 1, so the series is absolutely convergent by the ratio test.*0506

*That is the conclusion that we can make here.*0520

*Just to recap here, we look at the series that we are given,*0530

*We call that a _{n} .*0535

*Ratio test says you divide a _{n} +1/a_{n} .*0540

*aN+1 means you plug in n+1 wherever you saw n.*0544

*You did a little bit of algebra, the absolute values got rid of the negative signs, the factorials cancel each other out,*0548

*The limit came out to be 0, that is less than 1, and the ratio test then says that it is absolutely convergent. *0556

*The next example, we have n! over n ^{2} + 2.*0564

*Let us go for the ratio test again.*0568

*so a _{n} +1/a_{n} .*0573

*Is n+1!/n+1 ^{2} + 2/n!/^{2} + 2.*0578

*I am going to get rid of the absolute values because everything here is positive.*0596

*Again I am going to flip the denominator up and match it with the numerator.*0600

*We get n+1! divided by n! and then n ^{2} + 2/n+1^{2} + 2.*0604

*These are multiplied together.*0617

*n+1!/n!, remember you can write n+1! as n+1 × n!.*0621

*We learned that in a previous example so that cancels down to just n+1.*0634

*Now, n ^{2} + 2/n+1^{2} + 2.*0640

*This is a very common pattern where you have a polynomial.*0645

*The n version of it, in one part and the n+1 part in the other part.*0649

*The same pattern always works in that you have n ^{2} + some other stuff divided by,*0656

*Now if you work out n ^{2}, you can work it out, you are going to get n^{2} + some smaller terms, guaranteed.*0664

*If you divide top and bottom by n ^{2}, in the limit that thing is going to go to 1.*0673

*That always happens when you have a polynomial and you plug in n+1 and n.*0685

*It always goes to 1.*0692

*In that sense, the ratio test, what it is kind of doing, is sweeping a polynomial term away from our series because it is just sending that limit to 1.*0693

*What we have is n+1 × 1.*0705

*If we take the limit as n goes to infinity, of n+1 × 1, that is infinity which is certainly bigger than 1.*0710

*So l here, our l is infinity.*0722

*That is bigger than one.*0725

*The series diverges by the ratio test.*0729

*Remember in some of our other tests, like the limit comparison test,*0740

*You were not allowed to have infinite limits or 0 limits.*0747

*You were only allowed to have finite non-zero limits in the limit comparison test.*0752

*The ratio test is much more forgiving.*0754

*You can have any limit as long as the limit exists.*0758

*It can be 0, it can be any number, it can be infinity,*0762

*All you have to check is it less than 1 or is it bigger than 1.*0768

*If it turns out to be 1, that is the 1 case where you out of luck.*0772

*If it comes out to be one, then the ratio test does not give you an answer.*0776

*But 0 or infinity, they are ok for limits in the ratio test,*0780

*And they do give you an answer.*0783

*In this case, infinity is certainly bigger than 1 so we can say the series diverges by the ratio test.*0786

*Our third example is a _{n} = n+1^{n}/9^{n}.*0795

*That is one that is not very conducive to the ratio test.*0800

*The reason is that we have n's in the base and in the exponents.*0806

*I do not like using that for the ratio test.*0811

*Instead we are going to try the root test.*0814

*I am hoping that the root test will clear away some of those n's in the exponent.*0818

*Remember the root test says you look at a _{n} , well the nth root of that,*0829

*That is the same as a _{n} ^{1/n},*0832

*So that is n-1 ^{n}/9^{n}, absolute value of that to the ^{1/n},*0836

*The absolute values do not do anything here because the terms are positive.*0847

*n ^{1/n} is just 1, so we get n-1^{1}/9.*0855

*Then we take the limit of that.*0863

*As n goes to infinity, 9 is just a constant, so that limit is going to infinity.*0866

*Here, the l is infinity, which is certainly bigger than 1.*0873

*The series, we can say it diverges, by the root test, because we got a limit bigger than 1.*0880

*So, again, what made that series something that we wanted to look at using the root test.*0900

*The answer is that we had something with n in it, raised to the n power.*0910

*That is almost always a case that you want to refer to the root test.*0917

*When you see something with n in it, raised to the n power, that is the kind of thing that you want to use the root test for.*0925

*Because the ratio test is going to be really ugly.*0935

*Here we had an n-1 ^{n}, we know that we want to use the root test.*0936

*The root test says you look at a _{n} ^{1/n}, or the nth root of a_{n} ,*0944

*You work out the limit and again if it comes out to be bigger than 1 it diverges,*0950

*Less than 1 converges absolutely.*0955

*If you are unfortunate enough that it is equal to one then you have no information and the root test fails you.*0958

*So we will try some more examples later on.*0966

1 answer

Last reply by: Dr. William Murray

Thu Aug 4, 2016 6:03 PM

Post by Peter Ke on July 30, 2016

For example 1, how does e^n+1/e^n = e and how is (n+1)! the same as n!(n+1).

I know you explained it in the video but I still don't understand.

Please explain!

1 answer

Last reply by: Dr. William Murray

Thu Apr 4, 2013 5:45 PM

Post by Rohail Tariq on April 4, 2013

that was a brilliant video. i understood everything :)

1 answer

Last reply by: Dr. William Murray

Thu Apr 4, 2013 5:45 PM

Post by Ahmad Al-kheat on April 9, 2012

so just to recap here :)