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Lecture Comments (6)

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?

Alternating Series

Main theorems:

Alternating Series Test: Suppose bn > 0 for all n. Then ∑(−1)nbn (or ∑(−1)n+1bn ) is an alternating series. Check two conditions:

  1. Is bn decreasing? (i.e. is bn+1 < bn ?)

  2. Is lim bn = 0?

If both are true, then ∑(-1)nbn converges.

Estimates of sums: If an is a series that satisfies AST, and we use the partial sum sn as an estimate of the total sum S (the true answer), then our error E could be positive or negative, but it’s bounded by |an+1|:

|E| ≤ |an+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 (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 sn , or you can use it to reverse-engineer what n should be in order to know that a partial sum sn is accurate to within a specified error tolerance.

Alternating Series

Does the series ∑[(( − 1)n + 1)/(n + 1)] converge?
  • Check if an + 1< an
  • an + 1 = [1/(n + 2)],an = [1/(n + 1)]
  • [1/(n + 2)] < [1/(n + 1)]
  • an + 1< an
Find limn → ∞ anlimn → ∞ [1/(n + 1)] = 0 Thus the series converges
Does the series ∑[(2( − 1)n + 1)/(n − 3)] converge?
  • Check if an + 1< an
  • an + 1 = [2/(n − 2)],an = [2/(n − 3)]
  • [2/(n − 2)] < [2/(n − 3)]
  • an + 1< an
Find limn → ∞ anlimn → ∞ [2/(n − 3)] = 0 Thus the series converges
Does the series ∑[(( − 1)n + 1)/(n2)] converge?
  • Check if an + 1< an
  • an = [1/(n2)],an + 1 = [1/((n + 1)2)]
  • [1/((n + 1)2)] < [1/(n2)]
  • an + 1< an
Find limn → ∞ anlimn → ∞ [1/(n2)] = 0 Thus the series converges
Does the series ∑[(n3( − 1)n + 1)/(n2)] converge?
  • Check if an + 1< an
  • an = [(n3)/(n2)]
  • = n
n < [((n + 1)3)/((n + 1)2)]an + 1> an Thus the series diverges
Does the series [1/4] − [2/5] + [1/2] − [4/7] + ... + [(( − 1)n + 1n)/(n + 3)] converge?
  • Determine an + 1
  • an = [n/(n + 3)],an + 1 = [(n + 1)/(n + 4)]
a5 = [5/8] = 0.625a6 = [6/9] = 0.666...a6> a5, thus the series diverges
Does the series ∑[(( − 1)n + 1)/(3n)] converge?
  • Determine an + 1
  • an = [1/(3n)],an + 1 = [1/(3n + 1)]
  • Try an arbitrary positive integer for n, such as n = 1
  • a1 = [1/3] = 0.333...
  • a2 = [1/9] = 0.111....
  • a2< a1
Find limn → ∞ anlimn → ∞ [1/(3n)] = 0 Thus the series diverges
Does the series ∑[(( − 1)n + 1n!)/(2n)] converge?
  • Determine an + 1
  • an = [n!/(2n)],an + 1 = [(( n + 1 )!)/(2n + 1)]
a2 = [2/4] = 0.5a3 = [6/8] = 0.75a3> a2, thus the series diverges
Does the series ∑[(( − 1)n + 12n)/n!] converge?
  • Determine an + 1
  • an = [(2n)/n!],an + 1 = [(2n + 1)/(( n + 1 )!)]
  • Try an arbitrary positive integer for n, such as n = 2
  • a2 = [4/2]
  • a3 = [8/6]
  • a3< a2
Find limn → ∞ an using the Ratio Test
limn → → ∞ [(2n)/n!] = 0 Thus the series converges
Does the series ∑( − 1)n + 12( [1/3] )n − 1 converge?
r = − [1/3]r < 1 Thus the series converges
What is the sum of the ∑( − 1)n + 12( [1/3] )n − 1 ?
Apply the Sum equation
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.


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

Transcription: Alternating Series

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

We are given -1n+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 bn to be 1/n.0018

We do have that that is decreasing, bn+1 is less than bn.0025

We do have that the limit of bn=0.0033

Our two conditions are satisfied.0038

That means the alternating series converges.0040

So, the sum of -1n+1 bn,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 an+1.0079

By the first term that you cut off.0086

We want that to be less than 0.01.0090

We want an to be less than 0.01.0103

Well an+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 s99 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 -xn/n.0208

If you plug in x=-1, then we get ln(2)=the sum of -1n/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 -1n/en 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 bn, we just take everything except the -1n.0018

That is 1/en.0024

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

bN+1 should be less than bn.0030

That should not be an en that should be an en.0037

Certainly that is decreasing.0040

en+1 is bigger than en, so 1/en+1 is less than 1/en.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

-1n/en 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, an+1, if we take the absolute value, that knocks away any possible negative signs.0117

So this is 1/en < or = 0.001 is 1/1000.0124

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

So we get en > 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 en 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 an+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 210 = 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, e10 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 en+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 s6, 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 s6 for this series,0340

Add up terms and find the partial sum, you get s6 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 s6 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 s6 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 -1n, or -1n+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 bn 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 bn 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(an+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, -1n/n!.0200

Here the bn is just 1/n!.0205

We have to check our two conditions, whether the bn is decreasing, and whether the limit is 0.0210

If we look at bn+1 vs. bn, bn should be less than bn, 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 bn 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 -1n × sin(n)/n2.0287

So, I sort of strip off -1n.0294

We look at this part,0296

bn is going to be sin(n)/n2.0298

Now, right away we have a problem because remember I said that we want to have bn > 0.0306

Here, the n2, 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 bn 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 -1n/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

-1n/n!, the n = 0 term, gives us -10, that is 1.0407

0! by definition is 1.0413

The n=1 term is -11, 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 an + 1, the abs(an+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 ex is the sum of xn/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 -1n/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