WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today, we continue on our discussion of sequences and series with infinite geometric series.
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What are **infinite geometric series**? Well, this is a type of series in which there is an infinite number of terms.
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Earlier on, I mentioned that, for either arithmetic or geometric series, you may have a limited number of terms,
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which is a finite series; or it may go on indefinitely, which is an infinite series.
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For example, the geometric series 1 + 1/4 + 1/16 + 1/64 is a geometric series with 4 terms; and it has a common ratio r = 1/4.
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This is a finite series: it has a limited number of terms.
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Consider another geometric series: 3 + 6 + 12 + 24, and then, when you see the ellipses (the three dots),
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it indicates that it goes on indefinitely; so this is an infinite series.
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The sums, s < font size="-6" > n < /font > , are called partial sums of the infinite series.
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For example, I may decide that I want to find the first 7 terms of this series, or the first 5 terms of an infinite geometric series.
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That would just be a partial sum of this series.
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And we will talk, in a minute, about special cases, when you actually can find the sum overall of an infinite geometric series.
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So, recall from the previous lesson the formula for the sum of a geometric series,
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s < font size="-6" > n < /font > = the first term, times (1 - r^n), divided by (1 - r).
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So, if I were to try to find the first seven terms of this series, s₇, I could use this formula.
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I have the first term; I need to find r; recall that I can find r by taking a term and dividing it by the one that goes just before.
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So here, r = 2; so this is 3(1 - 2^n), so that is 7; and then, divided by 1 - 2.
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So here, I found the partial sum for this infinite geometric series.
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You can actually find the sum of an infinite geometric series (not just the partial sum) in some cases.
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And I know that this sounds counterintuitive--how can you find the sum of something that goes on forever?
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But if you look, you can see why.
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All right, first of all, it is very important to know that this is limited to cases in which the series is convergent.
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And the word "convergent" indicates that the sum converges toward a particular number.
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A series is convergent if and only if the absolute value of r is less than 1.
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So, if you are working with a geometric series in which the value of the common ratio is either greater than -1 or less than 1,
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such as 3/4, for example--if r is 3/4 (the absolute value of that is 3/4), or if r is -1/2 (I take the absolute value of that--it would be 1/2),
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both of these are convergent; I could actually find the sum of those.
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Consider the series 1/3 + 1/9 + 1/27, going on indefinitely.
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Remember, to find r (let's go up here and find r), we are going to take 1/9, divided by 1/3.
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This is the same as 1/9 times 3, which equals 3/9, which equals 1/3; therefore, r = 1/3.
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If I am asked to find the sum of this, I go ahead and use this formula, x equals the first term, which is 1/3, divided by 1 - 1/3, equals (1/3)/(2/3).
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1/3 divided by 2/3 is the same as 1/3 times 3/2, or 3/6.
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The sum of this infinite geometric series is 1/2.
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Let's look at this another way: just go ahead and add up some of the terms and see what happens.
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s₁ for this term is 1/3; that is all you have: 1/3.
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So, s₂ would be adding 1/3 + 1/9; 1/3 + 1/9 equals...giving this a common denominator,
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I multiply both the numerator and the denominator by 3 to get 3/9 + 1/9 is 4/9; that is s₂.
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s₃ = 1/3 + 1/9 + 1/27; well, I know that these two are equal to 4/9, so that is 4/9 + 1/27.
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So again, I need to get a common denominator; and if you work that out, you will find that s₃ is 13/27.
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s₄...I would continue on: 1/3 + 1/9 + 1/27 + 1/81...and if you figure that out, it becomes 40/81.
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I won't work out the rest of these right here; but I will just tell you that s₅ is 121/243.
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s₆ is 364/729; and s₇ is 1093/2187.
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Let's look at the pattern here: it started out as 1/3; then it became 4/9, 13/27, 40/81, 121/243, 364/729, 1093/2187.
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What is happening is: this sum is converging upon 1/2.
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Another way that we say this is that the limit is 1/2; and that is terminology you will hear later on in higher math courses.
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But for right now, just be aware that you can only find the sum of an infinite geometric series if it is convergent,
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meaning that, as you take more and more terms and add them to the series, add them, and get their sum,
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you will see that the sum of the series is converging upon a particular number.
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And so, we just use this formula as a great shortcut to find the sum.
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Sigma notation is something we discussed earlier on.
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Go back and look at the lectures on geometric sequences and geometric series, if any of these concepts are new.
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But sigma notation--you will recall that the Greek letter sigma means sum, and we used it with other geometric series.
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For an infinite series, you are going to have something like this.
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We have our lower index, i = 1; and this is the series as i goes from 1 to infinity.
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The difference here is that, instead of stopping at a specific value up here, it goes on through infinity.
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And again, we have the formula for the sequence right here.
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Another example, or a specific example, would be a series, again, where i goes from 1 to infinity,
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but we have a formula over here, 1/4 times 1/2, raised to the n - 1 power.
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So, what you could do, then, is put 1 in here and find your first term, a₁.
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Then, put 2 here; find your second term; and go on infinitely, because of the type of series that this is.
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We can use the concepts that we just learned to actually write a repeating decimal as a fraction.
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The sum formula I just described can be used in this way.
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A repeating decimal would be something like this: .44444...and it just goes on and on that way.
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What we do is rewrite this as a geometric series.
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1: First step--how do you do that?
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Well, look at what this really means: it really means 0.4 + 0.04 + 0.004 + 0.0004, and so on.
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So, I rewrote this, but just as a series; and it is an infinite geometric series.
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Then, find the sum of the series: recall that you can only find the sum of an infinite geometric series
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if the absolute value of r is less than 1; this is required, or you can't find the sum.
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Well, let's show here that we are OK, because the absolute value of r is actually less than 1.
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So, in order to find the common ratio, r, I am going to take .04, and I am going to divide it by the term before it, which is .4.
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So, I move the decimal over one place; that is going to give me .4/4, which is .1; so the common ratio, r, equals .1.
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.1 is less than 1, so I am fine: I can use the sum formula, the first term divided by (1 - r).
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So, the sum equals the first term, divided by (1 - .1), equals .4/.9.
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Dividing that, move the decimal over; you can get rid of that decimal; that gives me 4/9.
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So, I found that the sum of this...
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I started out; I have this decimal that is a repeating decimal (it goes on forever); but I saw that I could rewrite this as a series.
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So, this, therefore, is equivalent to the sum of the series.
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My next step was to figure out what the sum of this series is, and it is actually 4/9.
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Therefore, .444 repeating can be rewritten as 4/9; those are equivalent.
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Recall that you can have a repeating decimal that doesn't just have one number repeat and repeat; it could be multiple numbers.
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It could be .383838 repeating, and you could do the same thing.
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You could rewrite this as .38 + .0038 + .000038, and so on; and then repeat as above, by finding the sum of the series.
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So, this could also be used for repeating decimals where there is a longer repeat.
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Also recall that we can write these as .4 with a bar over it, or .38 with a bar over it--that is just a different notation.
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All right, let's find the sum of this infinite geometric series.
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First, though, I am going to verify that I actually can find the sum of this by figuring out what r is.
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So, I am going to take 24, the common ratio, divided by 32; one term divided by the previous one gives me the common ratio.
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This simplifies out to 3/4: 3/4 is less than 1, so yes, I can find the sum, using this formula.
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The formula: I need to use the first term: 32/(1 - 3/4) = 32/(1/4).
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We can rewrite this as 32 times 4; and 32 times 4...4 times 2 is 8; 4 times 3 is 120; so that gives me 128.
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The sum of this infinite geometric series is 128, and I was able to find that using this formula, because I had an absolute value of r that is less than 1.
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Write as a fraction: recall that this notation means the same thing as .36363636, and so on.
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The first step is to write this as a geometric series, so I am going to rewrite this as .36 + .0036 + .000036, and so on.
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And I am going to use my formula for the sum of an infinite geometric series, which is the first term, divided by (1 - r).
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What is r? Well, as usual, I can find the common ratio, r, by taking a term, .0036, and dividing by the previous term, .36.
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Move the decimal over two places to get .36/36; and that is going to give me .01.
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.01 is less than 1, so I can find the sum.
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Take the first term; divide it by (1 - .01); this gives me .36/.99.
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I can move the decimal over two places, to give me 36/99; simplify that, because there is a common factor of 3; this is actually 12/33.
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I can see, again, that I have another common factor of 3; so this is going to give me 4/11.
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Therefore, this 0.36 repeating decimal can be rewritten as 4/11; so I wrote this repeating decimal as a fraction.
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Find the sum for this infinite geometric series: before I proceed, I check that the absolute value of the common ratio is actually less than 1.
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10/12...to find the common ratio, divide that by the term that came just before, which is 5/4.
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And this is going to give me 10/12 times 4/5 equals 40/60; so this is going to give me 2/3.
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And since 2/3 is less than 1, I can find the sum of this series using my formula for the sum of an infinite geometric series.
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OK, the first term is 5/4; I am dividing that by 1 - 2/3 to get 5/4 divided by 1/3.
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This is the same as 5/4; and then I just take the inverse of 1/3, and 5/4 times the inverse of 1/3; that is 3, so this is going to give me 15/4.
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And you can keep this as an improper fraction, or write it as a mixed number.
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The sum of this infinite geometric series is 15/4.
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Again, I was able to find that because the common ratio had an absolute value that was less than 1.
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Write as a fraction: we have another decimal that is repeating--goes on infinitely: 0.99999, and so on.
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Start out by rewriting this as a geometric series.
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I have 0.9 + 0.09 + 0.009, and so on; it goes on infinitely.
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Find the common ratio, r: r =...I am going to take .09, and divide that by .9.
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Move the decimal over one place to get .9/9; therefore, the common ratio is .1.
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Next, I use my formula for the sum of an infinite geometric series, which is the first term, divided by 1 - r.
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And this is going to give me the first term, .9, divided by 1 - .1; that is .9/.9; this is simply 1.
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So, this asked me to write it as a fraction; and .999 = 1, or you could say 1/1.
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This is still a fraction, because you could just write it as 1/1.
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And it is counterintuitive, if you think ".999 repeating is actually 1"; it doesn't seem like it is, but it is actually correct.
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That concludes this lesson of Educator.com on infinite geometric series; thanks for visiting!