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

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Alternating Series

**Main theorems:**

**Alternating Series Test:** Suppose *b*_{n} > 0 for all *n*. Then ∑(−1)* ^{n}b_{n}* (or ∑(−1)

*) is an*

^{n+1}b_{n}__alternating series__. Check two conditions:

Is

*b*decreasing? (i.e. is_{n}*b*_{n+1}<*b*?)_{n}Is lim

*b*= 0?_{n}

If both are true, then ∑(-1)* ^{n}b_{n}* converges.

**Estimates of sums:** If ∑*a _{n}* is a series that satisfies AST, and we use the partial sum

*s*as an estimate of the total sum

_{n}*S*(the true answer), then our error

*E*could be positive or negative, but it’s bounded by |

*a*

_{n+1}|:

|*E*| ≤ |*a*_{n+1}|

**Hints and tips:**

You must have an alternating series for the Alternating Series Test to apply. Alternating series are usually flagged by (−1)

^{n}, (−1)^{n−1}, (−1)^{n+1}, or cos*nπ*(which works out to (−1)^{n}). (However, if you have two of these expressions, the negatives cancel and the series is not alternating.) If other parts of the expression are not reliably positive (like sin*n*, which is sometimes positive and sometimes negative in no particular pattern), then AST does not apply.The AST is a one-way test − you can only use it to show that a series converges. If its conditions are not met, you cannot say that the series diverges. The AST simply tells you nothing, and you must find another test.

You can only use the AST Error Estimate if the AST applies to the series.

There are other Error Estimates from other tests (the Integral Test, for example), but the AST Error Estimate is the easiest to use if it applies.

You can use the AST Error Estimate to determine the error when you add up a particular partial sum

*s*, or you can use it to reverse-engineer what_{n}*n*should be in order to know that a partial sum*s*is accurate to within a specified error tolerance._{n}

### Alternating Series

^{n + 1})/(n + 1)] converge?

- Check if a
_{n + 1}< a_{n} - a
_{n + 1}= [1/(n + 2)],a_{n}= [1/(n + 1)] - [1/(n + 2)] < [1/(n + 1)]
- a
_{n + 1}< a_{n}

_{n → ∞}a

_{n}lim

_{n → ∞}[1/(n + 1)] = 0 Thus the series converges

^{n + 1})/(n − 3)] converge?

- Check if a
_{n + 1}< a_{n} - a
_{n + 1}= [2/(n − 2)],a_{n}= [2/(n − 3)] - [2/(n − 2)] < [2/(n − 3)]
- a
_{n + 1}< a_{n}

_{n → ∞}a

_{n}lim

_{n → ∞}[2/(n − 3)] = 0 Thus the series converges

^{n + 1})/(n

^{2})] converge?

- Check if a
_{n + 1}< a_{n} - a
_{n}= [1/(n^{2})],a_{n + 1}= [1/((n + 1)^{2})] - [1/((n + 1)
^{2})] < [1/(n^{2})] - a
_{n + 1}< a_{n}

_{n → ∞}a

_{n}lim

_{n → ∞}[1/(n

^{2})] = 0 Thus the series converges

^{3}( − 1)

^{n + 1})/(n

^{2})] converge?

- Check if a
_{n + 1}< a_{n} - a
_{n}= [(n^{3})/(n^{2})] - = n

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

^{2})]a

_{n + 1}> a

_{n}Thus the series diverges

^{n + 1}n)/(n + 3)] converge?

- Determine a
_{n + 1} - a
_{n}= [n/(n + 3)],a_{n + 1}= [(n + 1)/(n + 4)]

_{5}= [5/8] = 0.625a

_{6}= [6/9] = 0.666...a

_{6}> a

_{5}, thus the series diverges

^{n + 1})/(3

^{n})] converge?

- Determine a
_{n + 1} - a
_{n}= [1/(3^{n})],a_{n + 1}= [1/(3^{n + 1})] - Try an arbitrary positive integer for n, such as n = 1
- a
_{1}= [1/3] = 0.333... - a
_{2}= [1/9] = 0.111.... - a
_{2}< a_{1}

_{n → ∞}a

_{n}lim

_{n → ∞}[1/(3

^{n})] = 0 Thus the series diverges

^{n + 1}n!)/(2

^{n})] converge?

- Determine a
_{n + 1} - a
_{n}= [n!/(2^{n})],a_{n + 1}= [(( n + 1 )!)/(2^{n + 1})]

_{2}= [2/4] = 0.5a

_{3}= [6/8] = 0.75a

_{3}> a

_{2}, thus the series diverges

^{n + 1}2

^{n})/n!] converge?

- Determine a
_{n + 1} - a
_{n}= [(2^{n})/n!],a_{n + 1}= [(2^{n + 1})/(( n + 1 )!)] - Try an arbitrary positive integer for n, such as n = 2
- a
_{2}= [4/2] - a
_{3}= [8/6] - a
_{3}< a_{2}

_{n → ∞}a

_{n}using the Ratio Test

lim

_{n → → ∞}[(2

^{n})/n!] = 0 Thus the series converges

^{n + 1}2( [1/3] )

^{n − 1}converge?

^{n + 1}2( [1/3] )

^{n − 1}?

Sum = [a/(1 − r)] = [2/(1 + [1/3])] = [2/([4/3])] = [6/4]

*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

### Alternating 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
- Main Theorems 0:05
- Alternation Series Test (Leibniz)
- How It Works
- Two Conditions
- Never Use for Divergence
- Estimates of Sums
- Lecture Example 1 3:19
- Lecture Example 2 4:46
- Lecture Example 3 6:28
- 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: Alternating Series

*I would like to try another example here of the alternating series test.*0000

*We are given -1 ^{n+1}/n.*0005

*We want to show that it converges and determine how many times would be necessary to estimate the sum within 0.01.*0010

*So, here let us take our b _{n} to be 1/n.*0018

*We do have that that is decreasing, b _{n}+1 is less than b_{n}.*0025

*We do have that the limit of b _{n}=0.*0033

*Our two conditions are satisfied.*0038

*That means the alternating series converges.*0040

*So, the sum of -1 ^{n+1} b_{n},*0045

*We can say that that certainly converges by the alternating series test.*0055

*That is the first part of what we had to do there.*0065

*We had to show that that series converged.*0070

*Then we have to determine how many terms are necessary to estimate the sum within 0.01.*0072

*Remember the error when you use the alternating series test, is bounded by a _{n}+1.*0079

*By the first term that you cut off.*0086

*We want that to be less than 0.01.*0090

*We want a _{n} to be less than 0.01.*0103

*Well a _{n}+1 in absolute value is 1/n+1.*0107

*That should be less than or equal to, 0.01 is the same as 1/100.*0113

*So if we take the reciprocal of that, remember that reverses the inequality.*0119

*So this n+1 > or = 100.*0124

*So n would have to be bigger than 99.*0130

*We would need 99 terms.*0138

*We would need to add up 99 terms of this series to get an estimate that is within 0.01 of the true answer.*0147

*In fact, on the computer earlier, I did add up 99 terms of these series.*0155

*It turns out that s _{99} is approximately equal to 0.698172.*0162

*If you actually add up all those 99 terms, that is the approximate answer that you get.*0173

*This series is another one that we can figure out using some techniques that we are going to learn later on.*0180

*Later, we are going to learn about Taylor Series.*0186

*I do not expect you to understand these yet because we have not developed the machinery for it yet.*0191

*But, later we will learn that the Taylor Series for the function ln(1-x) is the sum from n=1 to infinity,*0196

*Of -x ^{n}/n.*0208

*If you plug in x=-1, then we get ln(2)=the sum of -1 ^{n}/n,*0216

*Which is exactly the series we have been trying to estimate.*0237

*The true sum of the series is in fact ln(2).*0241

*That turns out to be approximately equal to 0.693147.*0247

*The true sum using techniques that we have not really developed the machinery for yet is 0.693147.*0257

*The sum we got was 0.698172, so in fact we were within 0.01 of the true sum.*0266

*Just to recap there,*0274

*What we were trying to show is how many terms do we need to estimate the true sum within 0.01.*0277

*Well, we have this error formula that says the error is bounded by the first term that we cut off, the n+1 term.*0284

*We set that less than 0.01, we would like the error to be less than 0.01,*0290

*Then we solve that out, get a value of n, and then that tells us how many terms we need to take to estimate our sum.*0298

*The last example I would like to look at is to show that the series -1 ^{n}/e^{n} converges.*0000

*To determine what partial sum we would need to estimate the true sum within 0.001, so very very accurate estimate there.*0008

*Here, for our b _{n}, we just take everything except the -1^{n}.*0018

*That is 1/e ^{n}.*0024

*Then we check our two conditions, one is that it is decreasing.*0026

*bN+1 should be less than b _{n}.*0030

*That should not be an e _{n} that should be an e^{n}.*0037

*Certainly that is decreasing.*0040

*e ^{n+1} is bigger than e^{n}, so 1/e^{n+1} is less than 1/e^{n}.*0042

*Certainly, the limit of that as n goes to infinity is equal to 0.*0049

*So, that certainly justified.*0058

*The series, we can say the alternating series,*0061

*-1 ^{n}/e^{n} converges by the alternating series test.*0068

*That is the first half of the work that we needed to do there.*0084

*Now we have to figure out what partial sum we would need to estimate the true sum to within 0.001.*0088

*Again, we are going to use our error estimation formula.*0097

*It says the error is less than whatever term we cut off.*0102

*The absolute value of whatever term we cut off.*0108

*We would like that to be less than 0.001.*0110

*So, a _{n}+1, if we take the absolute value, that knocks away any possible negative signs.*0117

*So this is 1/e ^{n} < or = 0.001 is 1/1000.*0124

*If we take reciprocals, remember that reverses the inequality.*0135

*So we get e ^{n} > or = to 1000.*0140

*Now there are 2 possible ways you can finish this.*0144

*If you do not have a calculator handy, if you are on a test or you are not allowed to use calculators,*0147

*Or for some reason you are trying to make a quick estimate without any mechanical tools,*0150

*We want to figure out what n is, is it safe to say that e ^{n} is bigger than 1000.*0157

*Here is what I know.*0165

*I know that e is about 2.7.*0166

*That is certainly bigger than 2, right?*0171

*Sorry, a small correction here, this is a _{n}+1,*0176

*So my exponents on the e's should be n+1's.*0180

*Now I am trying to make that bigger than 1000,*0187

*And if I do not have a calculator, I estimate that e is bigger than 2.*0192

*I know that 2 ^{10} = 1024.*0198

*One way I know that is that when people talk about a kilobyte, or 1K of memory, is that they really mean 1024 bytes.*0204

*So that is bigger than 1000.*0217

*So certainly, e ^{10} is bigger than 1000 because e is bigger than 2.*0219

*So, if n+1 = 10, then n = 9 is enough.*0227

*So that is just a rough estimation based on no calculator.*0235

*If we do not have a calculator handy.*0240

*We would get n = 9 is enough.*0246

*If you do have access to a calculator and you can make this more precise, we could actually solve for n.*0250

*So, if you do have a calculator,*0255

*We can take e ^{n+1} > 1000, we can solve for n.*0260

*We can take ln(both sides) and get n+1 > or = ln(1000).*0264

*So n should be > or = ln(1000)-1.*0271

*If you check on a calculator, ln(1000) = 6.9077.*0280

*So - 1, we get n > 5.9077.*0288

*But, since n is the number of terms we are taking, it has to be a whole number, so n=6 is enough.*0297

*So if you have access to a calculator, you would use n=6 there.*0310

*If you do not have access to a calculator, you probably would get a slightly rougher answer.*0314

*The conclusion there is that you would use s _{6}, the partial sum for n=6.*0322

*That is really the answer to the question that we were asked, the partial sum would have to be s6.*0330

*Let us go ahead and see if we can check out how close that actually gets us.*0335

*If you actually add up s _{6} for this series,*0340

*Add up terms and find the partial sum, you get s _{6} is approximately equal to 0.731725.*0347

*That is the partial sum if you just run this series from n=0 up to n=6.*0361

*You get this partial sum s _{6} approximately 0.731725.*0367

*That would be your estimate for the true sum of this series.*0375

*Now this series is another one where you can actually find the true sum using the different tools.*0380

*Because, notice that this series is actually a geometric series.*0387

*It is just the sum of -1/e raised to the n power.*0395

*We have a formula for the sum of a geometric series.*0404

*The sum, remember, is, I like to think of the formula in terms of words.*0407

*If you write the formula in terms of variables, you get slightly different formulas depending on whether you start at n=1 or n=1,*0416

*You will see in different books different version of this formula.*0426

*The version that always works is to write it in words.*0432

*First term over one minus the common ratio.*0434

*Here, the first term is n=0, you plug in n=0 and you get 1.*0439

*The common ratio there is 1/e, so 1-1/e,*0446

*Sorry, the common ratio is -1/e, so 1- that is 1+1/e.*0455

*If you multiply top and bottom by e, you get e/e+1.*0465

*That in turn you can plug into a calculator and get 0.731059.*0473

*So, remember we were trying to estimate this series to within 0.001.*0482

*Our estimate was 0.731725.*0487

*The true sum, which we found out using geometric series techniques was 0.731059.*0491

*So, in fact, we did get within 0.001 of the true sum as we wanted to.*0500

*Just to recap there, the first thing we had to do was show that it converged.*0507

*We did that by checking these two conditions to invoke the alternating series test and they both worked.*0511

*The second thing was to find how many sums we need to estimate this series within, estimate the true sum within a certain error tolerance.*0519

*We used our alternating series error formula and set it less than the error we are willing to allow,*0528

*Because we want to have less error than 0.001,*0535

*We solve that out for n.*0540

*Now, that could go different ways depending on whether you have to make a rough estimate without a calculator,*0541

*Or whether you can find an exact value with a calculator.*0548

*We got a value of n, n=6,*0551

*That was really all we were asked for.*0555

*Just to check it, we did add up the partial sum s _{6} to get this 0.731.*0558

*Because this happened to be a geometric series, we could also use the geometric series formula to get the true sum.*0566

*In fact, both answers are 0.731 so they are within 0.001 of each other.*0575

*That is the end of our section on alternating series, this is educator.com.*0582

*This is educator.com and we are here to talk about alternating series.*0000

*The main test that we are going to be using is called the alternating series test.*0005

*In some Calculus classes, this is called the Leibnitz alternating series test.*0011

*I will be referring to it as AST for short.*0015

*Sometimes you might see it as LAST for short.*0019

*The alternating series test works like this.*0023

*You start out with some positive terms, and then you make that into an alternating positive and negative series,*0025

*By either multiplying it by -1 ^{n}, or -1^{n+1}.*0033

*If you started positive, that turns it into a series that alternates positive and negative.*0038

*There are two conditions that you have to check.*0044

*You first check is b _{n} decreasing?*0047

*Meaning, are they getting smaller, in other words, is each subsequent term smaller than that previous term?*0052

*Then you also have to check if the limit of b _{n} is 0.*0056

*If both of those terms work, if both of those conditions are satisfied,*0060

*Then you can say that the series converges by the alternating series test.*0065

*By the way, you can never use AST to say a series diverges.*0073

*That is a very common mistake that Calculus students make.*0084

*But, you will not make it because you should know that this is a one-way test.*0089

*It can only tell you that something converges.*0094

*It can NEVER tell you that a series diverges.*0097

*So, if these conditions work then you say that it converges.*0100

*If they do not work, you are out of luck and you have got to find another test.*0104

*The second thing we are going to do with the alternating series test is estimates of sums.*0110

*If you have a series that satisfies the alternating series test,*0116

*And you want to estimate the sum of that series,*0120

*What you might do is add up the first few terms and say well that is my guess as the sum of the series.*0124

*In other words, you would use the partial sum as an estimate of the total sum that we will call S.*0131

*S is the true answer but we do not know what that is.*0137

*SN is the partial sum, is your estimate.*0140

*Then we ask, what is the possible error that you could make by making that estimate.*0144

*The area, we do not know, it could be positive, it could be negative.*0150

*But, the area is bounded by the abs(a _{n}+1).*0153

*That means that this is the first term that you did not include in your estimate, in your partial sum.*0158

*That you cut off.*0174

*So, you will be adding up terms of a series.*0179

*At some point you stop, you cut it off, and you say I am not going to look at any more terms.*0180

*What is my possible error by just taking the terms I have looked at so far?*0185

*The answer is, your error is the first term that you cut off,*0191

*So it is sort of the next term.*0194

*We will see some examples of that so that you get some practice.*0199

*First up is a series here, -1 ^{n}/n!.*0200

*Here the b _{n} is just 1/n!.*0205

*We have to check our two conditions, whether the b _{n} is decreasing, and whether the limit is 0.*0210

*If we look at b _{n}+1 vs. b_{n}, b_{n} should be less than b_{n}, well that is 1/n+1!.*0218

*Vs. 1/n!.*0228

*Certainly n+1! is much bigger than n!,*0230

*So n+1/n+1! is less than 1/n!,*0238

*So that condition is certainly satisfied.*0243

*The limit of the b _{n} is 1/n!.*0247

*That certainly equals 0.*0253

*So, condition 1 and condition 2 are both satisfied.*0255

*The series, when you make that series alternating by attaching a -1 to the n,*0260

*The series converges by the alternating series test.*0270

*So, let us try another example there.*0284

*Here we have -1 ^{n} × sin(n)/n^{2}.*0287

*So, I sort of strip off -1 ^{n}.*0294

*We look at this part,*0296

*b _{n} is going to be sin(n)/n^{2}.*0298

*Now, right away we have a problem because remember I said that we want to have b _{n} > 0.*0306

*Here, the n ^{2}, no problem there, that is always positive.*0315

*But, the sin(n) that kind of oscillates all over the place.*0320

*Remember, sin oscillates between -1 and 1,*0325

*And sin(n) is going to be bouncing all around between -1 and 1.*0329

*So this is not always positive.*0332

*One of the requirements, before you even think about the alternating series test,*0340

*Was that the b _{n} be positive.*0348

*That fails right here so the alternating series test does not apply, which does not necessarily mean that the series diverges.*0351

*It just means that we do not know based on any of the tools that we have so far.*0367

*We have no conclusion.*0375

*We do not know based on the tools that we have learned so far,*0377

*Whether this series converges or diverges.*0379

*You cannot use the alternating series test.*0384

*So, let us try another example.*0387

*The sum of -1 ^{n}/n!.*0391

*Here, we want to find the partial sum s4,*0394

*Then determine how far away that might be from the true sum.*0396

*Let us write out some terms here.*0403

*-1 ^{n}/n!, the n = 0 term, gives us -1^{0}, that is 1.*0407

*0! by definition is 1.*0413

*The n=1 term is -1 ^{1}, so -1, and 1 factorial is also 1.*0419

*n=2 gives us positive 1 again, and then 2! = 2.*0427

*n=3 gives -1/3! is 1/6.*0437

*n=4 gives us +1 over 4! is 24.*0443

*Then we will write one more term, n=5.*0449

*n=5 gives us -1/5!, which is 1/120.*0454

*Now s4 means that you go up to the n=4 term, so we are going to add that up.*0462

*s4=1 - 1 + 1/2 - 1/6 + 1/24.*0467

*The 1 - 1 cancel, and if we put everything in terms of 24ths here, that is 12-4+1/24.*0480

*That is 9/24.*0490

*Which is 3/8, which is 0.375.*0496

*That is s4, our first answer.*0502

*Then they say, what is the maximum possible error in using that as an estimate of the true sum.*0506

*The error, remember is bounded by a _{n} + 1, the abs(a_{n}+1).*0515

*That is the first term that you cut off.*0525

*The first term that we cut off is, well it was -1/120, but since we take the absolute value, that is 1/120.*0528

*The error is at most 1/120, if you are looking for a decimal approximation, I can tell you that is less than 1/100 which is 0.01.*0542

*We are accurate to 2 decimal places.*0554

*We are accurate there to at least 2 decimal places.*0567

*In fact, later we are going to be studying Taylor Series, so this is not something that you are supposed to understand yet.*0574

*We will find that the Taylor Series for e ^{x} is the sum of x^{n}/n!.*0582

*That is Taylor Series stuff, we have not learned it yet.*0595

*When we do learn it, we will realize that e ^{-1} is the sum of -1^{n}/n!.*0598

*Which is exactly the series we have been looking at.*0607

*And e ^{-1}, that is 1/e, if you plug that in, then you get approximately 0.3679.*0610

*That is really the true sum of the series, 0.3679.*0624

*What we got was 0.375, so we did get pretty close.*0629

*We did in fact get closer than 0.01, so our estimate is accurate.*0636

1 answer

Last reply by: Dr. William Murray

Thu Mar 27, 2014 6:42 PM

Post by Brandyn Albrecht on March 27, 2014

Why can we take the (-1)^n out when solving for bn? Wouldn't the sign of the answer change as n goes from an even to odd number on it's way to infinity? I would think it would be like the sine function where it switches from positive to negative depending on the n being used. I could see taking a (+1)^n out but a negative I feel like wouldn't work because it's not always positive. Similar to Example 2 not always being positive.

3 answers

Last reply by: Dr. William Murray

Mon Nov 5, 2012 10:03 PM

Post by Monica Khun on August 8, 2012

why can't this be watched?