### Infinite Geometric Series

- Remember that an infinite geometric series converges only if |r| < 1. If you are given a series to evaluate, first check the value of r. If it does not satisfy this condition, the series does not have a sum.
- To convert a repeating decimal to a fraction, let r = 10
^{-n}, where n = number of digits in the repeating pattern of the decimal. Let a_{1}= the fraction which contains the pattern written once in the numerator and 10^{n}in the denominator.

### Infinite Geometric Series

- Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a
_{1})/(1 − r)] - r = [(a
_{2})/(a_{1})] = [( − [4/5])/1] = − [4/5] - Step 2) Since r is < 1, you may find the sum using the formula
- S = [(a
_{1})/(1 − r)] - S = [(a
_{1})/(1 − r)] = [1/(1 − ( − [4/5] ))] = - [1/(1 + [4/5])] = [1/([9/5])] = [5/9]

- Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a
_{1})/(1 − r)] - r = [(a
_{2})/(a_{1})] = [( − [5/12])/([5/3])] = − [5/12] ÷[5/3] = − [5/12]*[3/5] = − [3/12] = − [1/4] - Step 2) Since r is < 1, you may find the sum using the formula
- S = [(a
_{1})/(1 − r)] - S = [(a
_{1})/(1 − r)] = [([5/3])/(1 − ( − [1/4] ))] = - [([5/3])/(1 − ( − [1/4] ))] = [([5/3])/(1 + [1/4])] = [([5/3])/([5/4])]
- [([5/3])/([5/4])] = [5/3]*[4/5] = [4/3]

- Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a
_{1})/(1 − r)] - r = [(a
_{2})/(a_{1})] = [3.52/4.4] = 0.8 - Step 2) Since r is < 1, you may find the sum using the formula
- S = [(a
_{1})/(1 − r)] - S = [(a
_{1})/(1 − r)] = [4.4/(1 − ( 0.8 ))] = - [4.4/(1 − ( 0.8 ))] = [4.4/0.2] = 22

- Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a
_{1})/(1 − r)] - r = [(a
_{2})/(a_{1})] = [1.86/9.3] = 0.2 - Step 2) Since r is < 1, you may find the sum using the formula
- S = [(a
_{1})/(1 − r)] - S = [(a
_{1})/(1 − r)] = [9.3/(1 − ( 0.2 ))] = - [9.3/(1 − ( 0.2 ))] = [9.3/0.8] = 11.625

- Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a
_{1})/(1 − r)] - r = [(a
_{2})/(a_{1})] = [([5/2])/5] = [5/2] ÷[5/1] = [5/2]*[1/5] = [1/2] - Step 2) Since r is < 1, you may find the sum using the formula
- S = [(a
_{1})/(1 − r)] - S = [(a
_{1})/(1 − r)] = [5/(1 − ( [1/2] ))] = - [5/(1 − ( [1/2] ))] = [5/([1/2])] = 10

- Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a
_{1})/(1 − r)] - r = [(a
_{2})/(a_{1})] = [([1/10])/([1/2])] = [1/10] ÷[1/2] = [1/10]*[2/1] = [1/5] - Step 2) Since r is < 1, you may find the sum using the formula
- S = [(a
_{1})/(1 − r)] - S = [(a
_{1})/(1 − r)] = [([1/2])/(1 − ( [1/5] ))] = - [([1/2])/(1 − ( [1/5] ))] = [([1/2])/([4/5])] =
- [([1/2])/([4/5])] = [1/2]*[5/4] = [5/8]

- Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a
_{1})/(1 − r)] - r = [(a
_{2})/(a_{1})] = [( − 2)/6] = − [1/3] - Step 2) Since r is < 1, you may find the sum using the formula
- S = [(a
_{1})/(1 − r)] - S = [(a
_{1})/(1 − r)] = [6/(1 − ( − [1/3] ))] = - [6/(1 − ( − [1/3] ))] = [6/([4/3])] =
- [6/([4/3])] = 6*[3/4] = [18/4] = [9/2]

- Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a
_{1})/(1 − r)] - r = [(a
_{2})/(a_{1})] = [( − [2/5])/( − 2)] = − [2/5]* − [1/2] = [1/5] - Step 2) Since r is < 1, you may find the sum using the formula
- S = [(a
_{1})/(1 − r)] - S = [(a
_{1})/(1 − r)] = [( − 2)/(1 − ( [1/5] ))] = - [( − 2)/(1 − ( [1/5] ))] = [( − 2)/([4/5])] =
- [( − 2)/([4/5])] = − 2*[5/4] = − [5/2]

- Step 1 - Write the decimal as a geometric series
- 0.25252525... = 0.25 + 0.0025 + 0.000025 + 0.000025...
- Step 2 - Find the common ratio r
- r = [(a
_{2})/(a_{1})] = [0.0025/0.25] = 0.01 - Step 3 - Use the formula S
_{n}= [(a_{1})/(1 − r)] - S
_{n}= [(a_{1})/(1 − r)] = [0.25/(1 − 0.01)] = [0.25/0.99] - Move the decimal two times over
- [0.25/0.99] = [25/99]

- Step 1 - Write the decimal as a geometric series
- 0.25252525... = 0.15 + 0.0015 + 0.000015 + 0.000015...
- Step 2 - Find the common ratio r
- r = [(a
_{2})/(a_{1})] = [0.0015/0.15] = 0.01 - Step 3 - Use the formula S
_{n}= [(a_{1})/(1 − r)] - S
_{n}= [(a_{1})/(1 − r)] = [0.15/(1 − 0.01)] = [0.15/0.99] - Move the decimal two times over
- [0.15/0.99] = [15/99]

*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

### Infinite Geometric 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
- What are Infinite Geometric Series 0:10
- Example: Finite
- Example: Infinite
- Partial Sums
- Formula
- Sum of an Infinite Geometric Series 2:39
- Convergent Series
- Example: Sum of Convergent Series
- Sigma Notation 7:31
- Example: Sigma
- Repeating Decimals 8:42
- Example: Repeating Decimal
- Example 1: Sum of Infinite Geometric Series 12:15
- Example 2: Repeating Decimal 13:24
- Example 3: Sum of Infinite Geometric Series 15:14
- Example 4: Repeating Decimal 16:48

### Algebra 2

### Transcription: Infinite Geometric Series

*Welcome to Educator.com.*0000

*Today, we continue on our discussion of sequences and series with infinite geometric series.*0002

*What are infinite geometric series? Well, this is a type of series in which there is an infinite number of terms.*0010

*Earlier on, I mentioned that, for either arithmetic or geometric series, you may have a limited number of terms,*0018

*which is a finite series; or it may go on indefinitely, which is an infinite series.*0024

*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.*0030

*This is a finite series: it has a limited number of terms.*0047

*Consider another geometric series: 3 + 6 + 12 + 24, and then, when you see the ellipses (the three dots),*0052

*it indicates that it goes on indefinitely; so this is an infinite series.*0063

*The sums, s _{n}, are called partial sums of the infinite series.*0070

*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.*0074

*That would just be a partial sum of this series.*0084

*And we will talk, in a minute, about special cases, when you actually can find the sum overall of an infinite geometric series.*0087

*So, recall from the previous lesson the formula for the sum of a geometric series,*0097

*s _{n} = the first term, times (1 - r^{n}), divided by (1 - r).*0103

*So, if I were to try to find the first seven terms of this series, s _{7}, I could use this formula.*0114

*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.*0120

*So here, r = 2; so this is 3(1 - 2 ^{n}), so that is 7; and then, divided by 1 - 2.*0130

*So here, I found the partial sum for this infinite geometric series.*0149

*You can actually find the sum of an infinite geometric series (not just the partial sum) in some cases.*0160

*And I know that this sounds counterintuitive--how can you find the sum of something that goes on forever?*0166

*But if you look, you can see why.*0172

*All right, first of all, it is very important to know that this is limited to cases in which the series is convergent.*0178

*And the word "convergent" indicates that the sum converges toward a particular number.*0184

*A series is convergent if and only if the absolute value of r is less than 1.*0191

*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,*0196

*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),*0205

*both of these are convergent; I could actually find the sum of those.*0219

*Consider the series 1/3 + 1/9 + 1/27, going on indefinitely.*0222

*Remember, to find r (let's go up here and find r), we are going to take 1/9, divided by 1/3.*0234

*This is the same as 1/9 times 3, which equals 3/9, which equals 1/3; therefore, r = 1/3.*0241

*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).*0258

*1/3 divided by 2/3 is the same as 1/3 times 3/2, or 3/6.*0273

*The sum of this infinite geometric series is 1/2.*0283

*Let's look at this another way: just go ahead and add up some of the terms and see what happens.*0288

*s _{1} for this term is 1/3; that is all you have: 1/3.*0297

*So, s _{2} would be adding 1/3 + 1/9; 1/3 + 1/9 equals...giving this a common denominator,*0304

*I multiply both the numerator and the denominator by 3 to get 3/9 + 1/9 is 4/9; that is s _{2}.*0313

*s _{3} = 1/3 + 1/9 + 1/27; well, I know that these two are equal to 4/9, so that is 4/9 + 1/27.*0322

*So again, I need to get a common denominator; and if you work that out, you will find that s _{3} is 13/27.*0333

*s _{4}...I would continue on: 1/3 + 1/9 + 1/27 + 1/81...and if you figure that out, it becomes 40/81.*0343

*I won't work out the rest of these right here; but I will just tell you that s _{5} is 121/243.*0360

*s _{6} is 364/729; and s_{7} is 1093/2187.*0370

*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.*0383

*What is happening is: this sum is converging upon 1/2.*0400

*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.*0414

*But for right now, just be aware that you can only find the sum of an infinite geometric series if it is convergent,*0422

*meaning that, as you take more and more terms and add them to the series, add them, and get their sum,*0428

*you will see that the sum of the series is converging upon a particular number.*0435

*And so, we just use this formula as a great shortcut to find the sum.*0440

*Sigma notation is something we discussed earlier on.*0452

*Go back and look at the lectures on geometric sequences and geometric series, if any of these concepts are new.*0456

*But sigma notation--you will recall that the Greek letter sigma means sum, and we used it with other geometric series.*0462

*For an infinite series, you are going to have something like this.*0470

*We have our lower index, i = 1; and this is the series as i goes from 1 to infinity.*0476

*The difference here is that, instead of stopping at a specific value up here, it goes on through infinity.*0484

*And again, we have the formula for the sequence right here.*0490

*Another example, or a specific example, would be a series, again, where i goes from 1 to infinity,*0494

*but we have a formula over here, 1/4 times 1/2, raised to the n - 1 power.*0501

*So, what you could do, then, is put 1 in here and find your first term, a _{1}.*0507

*Then, put 2 here; find your second term; and go on infinitely, because of the type of series that this is.*0513

*We can use the concepts that we just learned to actually write a repeating decimal as a fraction.*0523

*The sum formula I just described can be used in this way.*0529

*A repeating decimal would be something like this: .44444...and it just goes on and on that way.*0533

*What we do is rewrite this as a geometric series.*0542

*1: First step--how do you do that?*0547

*Well, look at what this really means: it really means 0.4 + 0.04 + 0.004 + 0.0004, and so on.*0556

*So, I rewrote this, but just as a series; and it is an infinite geometric series.*0573

*Then, find the sum of the series: recall that you can only find the sum of an infinite geometric series*0580

*if the absolute value of r is less than 1; this is required, or you can't find the sum.*0590

*Well, let's show here that we are OK, because the absolute value of r is actually less than 1.*0604

*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.*0609

*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.*0618

*.1 is less than 1, so I am fine: I can use the sum formula, the first term divided by (1 - r).*0626

*So, the sum equals the first term, divided by (1 - .1), equals .4/.9.*0635

*Dividing that, move the decimal over; you can get rid of that decimal; that gives me 4/9.*0645

*So, I found that the sum of this...*0652

*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.*0657

*So, this, therefore, is equivalent to the sum of the series.*0665

*My next step was to figure out what the sum of this series is, and it is actually 4/9.*0670

*Therefore, .444 repeating can be rewritten as 4/9; those are equivalent.*0674

*Recall that you can have a repeating decimal that doesn't just have one number repeat and repeat; it could be multiple numbers.*0683

*It could be .383838 repeating, and you could do the same thing.*0690

*You could rewrite this as .38 + .0038 + .000038, and so on; and then repeat as above, by finding the sum of the series.*0698

*So, this could also be used for repeating decimals where there is a longer repeat.*0717

*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.*0722

*All right, let's find the sum of this infinite geometric series.*0736

*First, though, I am going to verify that I actually can find the sum of this by figuring out what r is.*0741

*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.*0746

*This simplifies out to 3/4: 3/4 is less than 1, so yes, I can find the sum, using this formula.*0753

*The formula: I need to use the first term: 32/(1 - 3/4) = 32/(1/4).*0769

*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.*0778

*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.*0791

*Write as a fraction: recall that this notation means the same thing as .36363636, and so on.*0806

*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.*0821

*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).*0844

*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.*0853

*Move the decimal over two places to get .36/36; and that is going to give me .01.*0864

*.01 is less than 1, so I can find the sum.*0871

*Take the first term; divide it by (1 - .01); this gives me .36/.99.*0875

*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.*0885

*I can see, again, that I have another common factor of 3; so this is going to give me 4/11.*0896

*Therefore, this 0.36 repeating decimal can be rewritten as 4/11; so I wrote this repeating decimal as a fraction.*0901

*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.*0918

*10/12...to find the common ratio, divide that by the term that came just before, which is 5/4.*0927

*And this is going to give me 10/12 times 4/5 equals 40/60; so this is going to give me 2/3.*0933

*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.*0949

*OK, the first term is 5/4; I am dividing that by 1 - 2/3 to get 5/4 divided by 1/3.*0961

*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.*0977

*And you can keep this as an improper fraction, or write it as a mixed number.*0989

*The sum of this infinite geometric series is 15/4.*0993

*Again, I was able to find that because the common ratio had an absolute value that was less than 1.*0999

*Write as a fraction: we have another decimal that is repeating--goes on infinitely: 0.99999, and so on.*1009

*Start out by rewriting this as a geometric series.*1021

*I have 0.9 + 0.09 + 0.009, and so on; it goes on infinitely.*1027

*Find the common ratio, r: r =...I am going to take .09, and divide that by .9.*1041

*Move the decimal over one place to get .9/9; therefore, the common ratio is .1.*1048

*Next, I use my formula for the sum of an infinite geometric series, which is the first term, divided by 1 - r.*1055

*And this is going to give me the first term, .9, divided by 1 - .1; that is .9/.9; this is simply 1.*1063

*So, this asked me to write it as a fraction; and .999 = 1, or you could say 1/1.*1081

*This is still a fraction, because you could just write it as 1/1.*1090

*And it is counterintuitive, if you think ".999 repeating is actually 1"; it doesn't seem like it is, but it is actually correct.*1093

*That concludes this lesson of Educator.com on infinite geometric series; thanks for visiting!*1105

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