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## Integral Calculus

### Unit 5: Lesson 2

Infinite geometric series- Worked example: convergent geometric series
- Worked example: divergent geometric series
- Infinite geometric series
- Infinite geometric series word problem: bouncing ball
- Infinite geometric series word problem: repeating decimal
- Proof of infinite geometric series formula
- Convergent & divergent geometric series (with manipulation)

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# Infinite geometric series word problem: repeating decimal

AP.CALC:

LIM‑7 (EU)

, LIM‑7.A (LO)

, LIM‑7.A.3 (EK)

, LIM‑7.A.4 (EK)

See how we can write a repeating decimal as an infinite geometric series. Created by Sal Khan.

## Want to join the conversation?

- What is the trick he used at6:45(12 votes)
- Here is a link to the video where Sal explains the tricks for testing divisibilty by 3 (and also for 2, 4, 5, 6, 9, and 10):

https://www.khanacademy.org/math/arithmetic/factors-multiples/divisibility_tests/v/divisibility-tests-for-2--3--4--5--6--9--10

Also, make sure you check out the comments for tricks for testing divisibility by 8 and 7!(26 votes)

- What if you have a number like 0.76383838383838?

It's starts with 0.76, then repeats the 38. This is really stumping me in my homework. Having trouble finding the common ratio to use.(6 votes)- I don't use Sal's method, but find the following more practical. This is mathematically equivalent to what Sal does, but it combines a bunch of steps and eliminates the subtraction.

The example you gave is particularly difficult to do, so it is a good one.

Step 1: Multiply and divide by a sufficient power of 10 to move the decimal place to the first of the repeating digits:

= (1/100)*(100)(0.7638383....)

= (1/100)(76.38383....)

Step 2: split the number into whole number and decimal portions

= (1/100)(76+ 0.38383....)

Step 3: Multiply and divide by as many 9s as there are repeating digits. One repeating digit means multiply by 9, two repeating digits means multiply by 99, three repeating digits means multiply by 999, etc.

= (1/100)(1/99) [99(76+ 0.38383....)]

= (1/9900) [(99*76) + 99(0.38383...)]

This uses a mathmatical trick so that the repeating portion multiplied by 9s will equal one set of the repeating digits. So 99(0.383838...) = 38

= (1/9900) [(99*76) + 38]

= (1/9900) [7525 + 38]

= (1/9900) [7525 + 38]

= (1/9900) [7562]

= 7562 / 9900

This is the repeating decimal as a fraction.

Step 4: Simplify if possible (not always easy). The GCF of 7562 and 9900 is 2, Thus

= 3781/4950

So, your example as a fraction in simplest form is 3781/4950.(15 votes)

- According to Sal's method, any repeating decimal can be expressed as an infinite geometric series with
`r = 0.1 or 0.01 or 0.001 or 0.0001 or so on`

. Is it safe for us to conclude that any repeating decimal can be expressed as a fraction with the denominator with this format`9 or 99 or 999 or 9999 or so on`

?(8 votes)- I guess it depends on what you mean by a "repeating decimal." If the decimal consists completely of a repeating sequence, then I'd say the answer is yes, since your r values will be as you stated and the first term can be taken as the repeating sequence.

However, what if you have a decimal such as 0.12323232323232323..., where just the 23 repeats? Would you consider this a repeating decimal? Probably so (and it's definitely a rational number), but it can't be strictly represented as a geometric sequence. The 0.1 at the start throws it off a bit. So it ends up looking like 122/990 in fractional form, which doesn't follow your pattern of 9, 99, 999, etc. What do you think?(10 votes)

- So, is that actually a proof that 9.9999... = 10?

I could write it as sum 9*(0.1)^k, from k = 0 to inf, which would result in

9/(1-0.1) = 10(7 votes)- 10-9.999... gives an answer that is closer to 0 by any other number you can give - this is why the two numbers are equal (allthough written in different ways)(3 votes)

- @4:00. Why not put the k back into 10^-4 = .4008(10^-4k). Or am I missing something entirely.(4 votes)
- They are the same. (10^-4)^k = (10^-4k) = 1/(10^4k)(1 vote)

- What is .016 with the 16 repeating as a fraction in simplest form?(2 votes)
- Okay: Let x = .0161616.... Then 100x = 1.6161616.... What would 100x - x =?

100x -x = 99x = (1.6161616.... - .0161616....) = 1.6 Therefore x = 1.6/99 = 16/990 = 8/495

I can't think of any endlessly repeating real number that can't be expressed as a fraction by this method.

P.S. 1.6/99 = 16/990 by multiplying by 10/10. 16/990 = (2*8)/(2*495) = 8/495 by noting 2/2 = 1(3 votes)

- Doesn't |r| have to be less than 1? In the video, it is equal to 10.(1 vote)
- It is not equal to 10, it is equal to 10^-4, which is the same as 0.0001, which is less than 1.(5 votes)

- Why does Sal do anything with the denominator at5:22? I dont understand why he would expand those like that. It certainly gives a different answer if you do that. I was going through the questions and didn't use fractions like that but kept the decimals and my answer was 1.96 instead of 196. Can someone explain this to me?(2 votes)
- From his example: 1 - 0.0001 = 0.9999, which is also equal to (9999/10,000). He probably just does this to remove decimals, since 0.4008/(9,999/10,000) = (0.4008*10,000)/9,999 = 4,008/9,999. This is still equal to 0.4008/0.9999.(2 votes)

- This seems to obviously generalize to any repeating decimal, and so all repeating decimals are rational - because they are all equal to the ratio of two integers. No?(1 vote)
- Yep! You're on point. Any repeating decimal can be represented as a/b, where a and b are integers. (with nonzero b)(4 votes)

- I got the sum from k=1 to infinity (4008/10^4k)

Is this equivalent to sal's answer?

Should I write it from k=0 to infinity because then i can check it's sum using a/1-r?

EDIT

My reasoning is because when k = 1, 4008/10^4 = .4008

k = 2, 4008/10^4(2) = .00004008

using inductive reasoning,

.400840084008....(2 votes)

## Video transcript

Let's say we have the
repeating decimal 0.4008, where the digits 4008
keep on repeating. So if we were to
write it out, it would look something like this. 0.400840084008, and it
keeps on going forever. What I want you to do right
now is pause the video and think about whether
you can represent this repeating decimal
as an infinite sum, as an infinite series. And then think about
whether that infinite series is a geometric series. So I'm assuming you've
given a go at it. So let's think about it. So for each term of
my infinite series, I'm going to represent one
of these repeating patterns of 4008. So, for example, I will make
this 4008 my first term. So this could be viewed as 0. And this 4008 represents 0.4008. Then I could make this
4008 my next term, or my next term will
represent this 4008. And this 4008 is the
same thing as 0.00004008. And then this next
4008, well, that represents 0 point-- and
we have eight zeroes-- 1, 2, 3, 4, 5, 6,
7, 8, 4008, 4008. And then we would just keep
on going like that forever. So we're just going to keep
on going like that forever. So hopefully there's
a pattern here. We're essentially throwing
four zeroes before the decimal every time. And we can just keep on
going like that forever. So this is an infinite sum. It's an infinite series. The next question is, is
this a geometric series? Well, in order for it to
be a geometric series, to go from one term
to the next, you must be multiplying
by the same value, by the same common ratio. So what are we multiplying when
we go from 0.4008 to this one right over here, where we add
four zeros before the 4008? What are we multiplying? Well, we moved the decimal
four spots to the left. So we're multiplying by
10 to the negative fourth. Or you could view it
as we're multiplying by 0.0001, 10 to the
negative 1, 2, 3, 4. To go from here to
here, well, same thing. Move the decimal four
places to the left. So once again, we're
multiplying by 0.0001. And so it looks
pretty clear that we have a common ratio of 10 to
the negative fourth power. So we can rewrite
all of this business as 0.4008 times our common
ratio for this first term times our common ratio of
10 to the negative fourth to the 0-th power--
so that gives us that right over there--
plus 0.4008 times 10 to the negative fourth
to the first power. And that gives us that
value right over there. Plus 0.4008 times 10 to the
negative 4 to the second power. And we keep on going. And so in this form, it
looks a little bit clearer, like a geometric series, an
infinite geometric series. And if we wanted to write
that out with sigma notation, we could write this
as the sum from k equals 0 to infinity
of, well, what's our first term going to be? It's going to be 0.4008
times our common ratio, which we could write out as either
10 to the negative fourth or 0.0001. I'll just write it as 10
to the negative fourth. 10 to the negative fourth to the
k-th power, to the k-th power. So the next
interesting question-- this clearly can be represented
as a geometric series-- is, well, what is the sum? You might say, well,
that's just going to be 4008 repeating
over and over. But I want to express
it as a fraction. And so I want you
to pause the video. Use what would you
already know about finding the sum of an infinite
geometric series to try to express this thing
right over here as a fraction. So I'm assuming
you've had a go at it. So let's think about it. We've already
seen, we've already derived in previous
videos, that the sum of an infinite
geometric series-- let me do this in
a neutral color. If I have a series like this,
k equals 0 to infinity of ar to the k power, that this sum
is going to be equal to a over 1 minus r. We've derived this actually
in several other videos. So in this case, this
is going to be-- well, our a here is 0.4008. And it's going to be that over 1
minus our common ratio, minus-- and I'll write it like this--
0.0001, 1 ten thousandth. So what's this going to be? Well, this is going to be
the same thing as 0.4008. If you take 1 minus
1 ten thousandths, or you could do this as
10,000 ten thousandths minus 1 ten thousandth, you're going
to have 9,999 ten thousandths. Once again, you could
view-- let me write this out just so this doesn't
look confusing. 1 is the same thing
as 10,000/10,000. And you're subtracting 1/10,000. And so you're going
to get 9,999/10,000. And so this is going
to be the same thing as 0.4008 times 10,000. So times 10,000 over 9,999. Well, what's this top
number times 10,000? Well that's just going
to give us 4,008. 4,008 over 9,999. And we've just expressed
that repeating decimal as a fraction. So we have succeeded. And you might say, well, maybe
we can simplify this thing. And so let's see. This is already a
fraction, so we've already kind of achieved it. But if we want to get
a little bit simpler. Let's see, if we add the
digits up here, 4 plus 8 is 12 and 1 plus 2 is 3. So this up here
is divisible by 3. And this down here is
clearly divisible by 3. So let's divide
both of them by 3. So 3 goes into
4,008-- let's see. It goes into 4 one time,
subtract, you get a 10. 3 times 3 is 9. Subtract, you get another 10. Goes into 3 times. 3 times 3 is 9. Subtract. Bring down an 8. 3 into 18 exactly six times. So our numerator is 1,336. This is no longer
divisible by 3. The sum of the digits
is not divisible by 3, it's not a multiple of 3. And if you divide this bottom
number by 3, you get 3,333. And I think we
have simplified it. I think we can simplified
it about as well as we can. Well, we could check more. Let me know if I didn't. But either way, we
have now written this. This was pretty neat. We saw that a
repeating decimal can be represented not just
as an infinite series, but as an infinite
geometric series. And then we were able to use
the formula that we derived for the sum of an
infinite geometric series to actually express
it as a fraction.