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INSTRUCTORSCarleen EatonGrant Fraser
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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 a1 = the fraction which contains the pattern written once in the numerator and 10n in the denominator.

Infinite Geometric Series

Find the sum 1 − [4/5] + [16/25] − [64/125]
  • Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a1)/(1 − r)]
  • r = [(a2)/(a1)] = [( − [4/5])/1] = − [4/5]
  • Step 2) Since r is < 1, you may find the sum using the formula
  • S = [(a1)/(1 − r)]
  • S = [(a1)/(1 − r)] = [1/(1 − ( − [4/5] ))] =
  • [1/(1 + [4/5])] = [1/([9/5])] = [5/9]
S = [5/9]
Find the sum [5/3] − [5/12] + [5/48] − [5/192]
  • Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a1)/(1 − r)]
  • r = [(a2)/(a1)] = [( − [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 = [(a1)/(1 − r)]
  • S = [(a1)/(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]
S = [4/3]
Find the sum 4.4 + 3.52 + 2.816 + 2.2528
  • Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a1)/(1 − r)]
  • r = [(a2)/(a1)] = [3.52/4.4] = 0.8
  • Step 2) Since r is < 1, you may find the sum using the formula
  • S = [(a1)/(1 − r)]
  • S = [(a1)/(1 − r)] = [4.4/(1 − ( 0.8 ))] =
  • [4.4/(1 − ( 0.8 ))] = [4.4/0.2] = 22
S = 22
Find the sum 9.3 + 1.86 + 0.372 + 0.0744...
  • Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a1)/(1 − r)]
  • r = [(a2)/(a1)] = [1.86/9.3] = 0.2
  • Step 2) Since r is < 1, you may find the sum using the formula
  • S = [(a1)/(1 − r)]
  • S = [(a1)/(1 − r)] = [9.3/(1 − ( 0.2 ))] =
  • [9.3/(1 − ( 0.2 ))] = [9.3/0.8] = 11.625
S = 11.625
Find the sum 5 + [5/2] + [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 = [(a1)/(1 − r)]
  • r = [(a2)/(a1)] = [([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 = [(a1)/(1 − r)]
  • S = [(a1)/(1 − r)] = [5/(1 − ( [1/2] ))] =
  • [5/(1 − ( [1/2] ))] = [5/([1/2])] = 10
S = 10
Find the sum [1/2] + [1/10] + [1/50] + [1/250]
  • Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a1)/(1 − r)]
  • r = [(a2)/(a1)] = [([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 = [(a1)/(1 − r)]
  • S = [(a1)/(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]
S = [5/8]
Find the sum 6 − 2 + [2/3] − [2/9]
  • Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a1)/(1 − r)]
  • r = [(a2)/(a1)] = [( − 2)/6] = − [1/3]
  • Step 2) Since r is < 1, you may find the sum using the formula
  • S = [(a1)/(1 − r)]
  • S = [(a1)/(1 − r)] = [6/(1 − ( − [1/3] ))] =
  • [6/(1 − ( − [1/3] ))] = [6/([4/3])] =
  • [6/([4/3])] = 6*[3/4] = [18/4] = [9/2]
S = [9/2]
Find the sum − 2 − [2/5] − [2/25] − [2/125]...
  • Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a1)/(1 − r)]
  • r = [(a2)/(a1)] = [( − [2/5])/( − 2)] = − [2/5]* − [1/2] = [1/5]
  • Step 2) Since r is < 1, you may find the sum using the formula
  • S = [(a1)/(1 − r)]
  • S = [(a1)/(1 − r)] = [( − 2)/(1 − ( [1/5] ))] =
  • [( − 2)/(1 − ( [1/5] ))] = [( − 2)/([4/5])] =
  • [( − 2)/([4/5])] = − 2*[5/4] = − [5/2]
S = − [5/2]
Write as a fraction 0.25252525....
  • 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 = [(a2)/(a1)] = [0.0025/0.25] = 0.01
  • Step 3 - Use the formula Sn = [(a1)/(1 − r)]
  • Sn = [(a1)/(1 − r)] = [0.25/(1 − 0.01)] = [0.25/0.99]
  • Move the decimal two times over
  • [0.25/0.99] = [25/99]
0.25252525... = [25/99]
Write as a fraction 0.15151515....
  • 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 = [(a2)/(a1)] = [0.0015/0.15] = 0.01
  • Step 3 - Use the formula Sn = [(a1)/(1 − r)]
  • Sn = [(a1)/(1 − r)] = [0.15/(1 − 0.01)] = [0.15/0.99]
  • Move the decimal two times over
  • [0.15/0.99] = [15/99]
0.15151515... = [15/99] = [5/33]

*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

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, sn, 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

sn = the first term, times (1 - rn), divided by (1 - r).0103

So, if I were to try to find the first seven terms of this series, s7, 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 - 2n), 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

s1 for this term is 1/3; that is all you have: 1/3.0297

So, s2 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 s2.0313

s3 = 1/3 + 1/9 + 1/27; well, I know that these two are equal to 4/9, so that is 4/9 + 1/27.0322

So again, I need to get a common denominator; and if you work that out, you will find that s3 is 13/27.0333

s4...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 s5 is 121/243.0360

s6 is 364/729; and s7 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, a1.0507

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 series0580

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