WEBVTT mathematics/calculus-ii/murray
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This is educator.com and we are here to talk about alternating series.
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The main test that we are going to be using is called the alternating series test.
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In some Calculus classes, this is called the Leibnitz alternating series test.
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I will be referring to it as AST for short.
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Sometimes you might see it as LAST for short.
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The alternating series test works like this.
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You start out with some positive terms, and then you make that into an alternating positive and negative series,
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By either multiplying it by -1^n, or -1^n+1.
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If you started positive, that turns it into a series that alternates positive and negative.
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There are two conditions that you have to check.
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You first check is b < font size="-6" > n < /font > decreasing?
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Meaning, are they getting smaller, in other words, is each subsequent term smaller than that previous term?
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Then you also have to check if the limit of b < font size="-6" > n < /font > is 0.
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If both of those terms work, if both of those conditions are satisfied,
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Then you can say that the series converges by the alternating series test.
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By the way, you can never use AST to say a series diverges.
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That is a very common mistake that Calculus students make.
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But, you will not make it because you should know that this is a one-way test.
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It can only tell you that something converges.
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It can NEVER tell you that a series diverges.
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So, if these conditions work then you say that it converges.
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If they do not work, you are out of luck and you have got to find another test.
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The second thing we are going to do with the alternating series test is estimates of sums.
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If you have a series that satisfies the alternating series test,
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And you want to estimate the sum of that series,
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What you might do is add up the first few terms and say well that is my guess as the sum of the series.
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In other words, you would use the partial sum as an estimate of the total sum that we will call S.
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S is the true answer but we do not know what that is.
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SN is the partial sum, is your estimate.
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Then we ask, what is the possible error that you could make by making that estimate.
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The area, we do not know, it could be positive, it could be negative.
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But, the area is bounded by the abs(a < font size="-6" > n < /font > +1).
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That means that this is the first term that you did not include in your estimate, in your partial sum.
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That you cut off.
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So, you will be adding up terms of a series.
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At some point you stop, you cut it off, and you say I am not going to look at any more terms.
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What is my possible error by just taking the terms I have looked at so far?
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The answer is, your error is the first term that you cut off,
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So it is sort of the next term.
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We will see some examples of that so that you get some practice.
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First up is a series here, -1^n/n!.
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Here the b < font size="-6" > n < /font > is just 1/n!.
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We have to check our two conditions, whether the b < font size="-6" > n < /font > is decreasing, and whether the limit is 0.
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If we look at b < font size="-6" > n < /font > +1 vs. b < font size="-6" > n < /font > , b < font size="-6" > n < /font > should be less than b < font size="-6" > n < /font > , well that is 1/n+1!.
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Vs. 1/n!.
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Certainly n+1! is much bigger than n!,
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So n+1/n+1! is less than 1/n!,
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So that condition is certainly satisfied.
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The limit of the b < font size="-6" > n < /font > is 1/n!.
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That certainly equals 0.
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So, condition 1 and condition 2 are both satisfied.
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The series, when you make that series alternating by attaching a -1 to the n,
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The series converges by the alternating series test.
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So, let us try another example there.
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Here we have -1^n × sin(n)/n².
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So, I sort of strip off -1^n.
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We look at this part,
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b < font size="-6" > n < /font > is going to be sin(n)/n².
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Now, right away we have a problem because remember I said that we want to have b < font size="-6" > n < /font > > 0.
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Here, the n², no problem there, that is always positive.
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But, the sin(n) that kind of oscillates all over the place.
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Remember, sin oscillates between -1 and 1,
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And sin(n) is going to be bouncing all around between -1 and 1.
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So this is not always positive.
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One of the requirements, before you even think about the alternating series test,
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Was that the b < font size="-6" > n < /font > be positive.
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That fails right here so the alternating series test does not apply, which does not necessarily mean that the series diverges.
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It just means that we do not know based on any of the tools that we have so far.
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We have no conclusion.
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We do not know based on the tools that we have learned so far,
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Whether this series converges or diverges.
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You cannot use the alternating series test.
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So, let us try another example.
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The sum of -1^n/n!.
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Here, we want to find the partial sum s4,
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Then determine how far away that might be from the true sum.
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Let us write out some terms here.
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-1^n/n!, the n = 0 term, gives us -1⁰, that is 1.
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0! by definition is 1.
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The n=1 term is -1¹, so -1, and 1 factorial is also 1.
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n=2 gives us positive 1 again, and then 2! = 2.
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n=3 gives -1/3! is 1/6.
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n=4 gives us +1 over 4! is 24.
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Then we will write one more term, n=5.
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n=5 gives us -1/5!, which is 1/120.
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Now s4 means that you go up to the n=4 term, so we are going to add that up.
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s4=1 - 1 + 1/2 - 1/6 + 1/24.
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The 1 - 1 cancel, and if we put everything in terms of 24ths here, that is 12-4+1/24.
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That is 9/24.
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Which is 3/8, which is 0.375.
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That is s4, our first answer.
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Then they say, what is the maximum possible error in using that as an estimate of the true sum.
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The error, remember is bounded by a < font size="-6" > n < /font > + 1, the abs(a < font size="-6" > n < /font > +1).
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That is the first term that you cut off.
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The first term that we cut off is, well it was -1/120, but since we take the absolute value, that is 1/120.
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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.
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We are accurate to 2 decimal places.
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We are accurate there to at least 2 decimal places.
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In fact, later we are going to be studying Taylor Series, so this is not something that you are supposed to understand yet.
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We will find that the Taylor Series for e^x is the sum of x^n/n!.
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That is Taylor Series stuff, we have not learned it yet.
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When we do learn it, we will realize that e^-1 is the sum of -1^n/n!.
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Which is exactly the series we have been looking at.
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And e^-1, that is 1/e, if you plug that in, then you get approximately 0.3679.
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That is really the true sum of the series, 0.3679.
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What we got was 0.375, so we did get pretty close.
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We did in fact get closer than 0.01, so our estimate is accurate.