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### Course: Calculus, all content (2017 edition) > Unit 7

Lesson 6: Infinite geometric series- Worked example: convergent geometric series
- Worked example: divergent geometric series
- Infinite geometric series
- Infinite geometric series formula intuition
- Proof of infinite geometric series as a limit
- Infinite geometric series word problem: bouncing ball
- Infinite geometric series word problem: repeating decimal
- Convergent & divergent geometric series (with manipulation)

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# Proof of infinite geometric series as a limit

The video provides a proof for the sum of an infinite geometric series using limits. When the absolute value of the common ratio (r) is between 0 and 1, the limit of the series converges to a finite sum. The formula for the sum is a / (1 - r), where a is the first term. Created by Sal Khan.

## Want to join the conversation?

- What would happen if the absolute value of r wasn't between 0 and 1? Would the sum just be infinite?(21 votes)
- Yes, the limit would be either positive or negative infinity. Or, as is usually stated in these cases we would say that "the series does not converge".(52 votes)

- So this equation only works when 'r' is between zero and one?(11 votes)
- Actually |r|<1 i.e -1<r<1. This the required condition for the equation to work. Hope it helped(22 votes)

- What does Zeno's paradox has to do with this?(8 votes)
- One of Zeno's paradoxes says that an arrow cannot reach its target because it first travels half the distance, then half the remaining distance, then half the remaining distance...

The total distance the arrow goes can be represented by a geometric series:

1/2 + 1/4 + 1/8 + 1/16 + ... = ∑ (1/2)^n from n=1 to oo (infinity)

As the geometric series approaches an infinite number of terms, the sum approaches 1.

What does this mean? The arrow of the paradox ultimately reaches its target. It takes an infinite number of steps to do it, but each step is also shorter. Thus you can piece the arrow's path into an infinite number of sections, but it is still going to complete its path in a finite time.(33 votes)

- How can you find a common 'r' if the first number in the series is zero?(4 votes)
- A geometric series cannot have it's first term be 0, since all other numbers of the series are created by multiplying the first term by the common ratio, and anything multiplied by 0 would continue to be 0, so the series would be something like: 0+0+0+0+0+...+0

If for some reason you discover that a geometric series has a first term equal to 0, then you can easily say that the total sum will also be 0.(7 votes)

- Hello, i need help with finding the sum of a geometric series i would appreciate it very much if you help, since i am not able to find the ratio i cant find the sum. (5/2^n - 1/3^n) with Sigma that has an index of n=0 and tends to infinity..(1 vote)
- The sum of this series is equal to the difference of the series 5/2^n and 1/3^n, the first of which is divergent and goes to infinity.(3 votes)

- 2:10, if
`|r| > 1`

, the number will become massively huge. How can we come up with the formula`Σ = a / (1 - r) if |r| > 1`

? Please notice that I fully understand`Σ = a / (1 - r) if 0 < |r| < 1`

. Did I miss something there?(5 votes)- You're right--the formula Σ = a / (1 - r) does not apply if |r| > 1. So our formula for the sum of a geometric series only applies if |r| < 1, which Sal begins to address around2:30.(3 votes)

- I thought the formula is a(1-r^n)/1-r

why did it expands to n+1 ? (a-ar^n+1)/1-r

?(5 votes) - How do you find the ratio if it isn't given?(2 votes)
- See what you have to multiply to go from one term to the next and that's the ratio. If it's not obvious to you, you can find the first term and the second term. Divide the second term by the first term and that's your ratio.(4 votes)

- Is there any difference between value and absolute value?(1 vote)
- A value can be positive or negative. An absolute value is always positive.(6 votes)

- At2:10he says that if r=1 denominator is 0, and we can't divide by zero. But, in that case numerator would also be 0, since a-a*(1)^n=0. Isn't lim 0/0=1?(2 votes)
- No, the limit of 0/0 is undefined, and since the limit is for the variable n and not for r, you cannot use any of the limit techniques to get rid of the 0/0.(3 votes)

## Video transcript

In a previous video,
we derived the formula for the sum of a
finite geometric series where a is the first term
and r is our common ratio. What I want to do
in this video is now think about the sum of an
infinite geometric series. And I've always found this
mildly mind blowing because, or actually more than mildly
mind blowing, because you're taking the sum of an infinite
things but as we see, you can actually
get a finite value depending on what
your common ratio is. So there's a couple of
ways to think about it. One is, you could
say that the sum of an infinite
geometric series is just a limit of this as n
approaches infinity. So we could say,
what is the limit as n approaches infinity of
this business, of the sum from k equals zero to n of
a times r to the k. Which would be the same
thing as taking the limit as n approaches infinity
right over here. So that would be the
same thing as the limit as n approaches infinity
of all of this business. Let me just copy
and paste that so I don't have to keep
switching colors. So copy and then paste. So what's the limit as n
approaches infinity here? Let's think about
that for a second. I encourage you to
pause the video, and I'll give you one hint. Think about it for r
is greater than one, for r is equal to
one, and actually let me make it
clear-- let's think about it for the absolute
values of r is greater than one, the absolute values
of r equal to one, and then the absolute
value of r less than one. Well, I'm assuming
you've given a go at it. So if the absolute value
of r is greater than one, as this exponent explodes,
as it approaches infinity, this number is just going
to become massively, massively huge. And so the whole thing
is just going to become, or at least you could
think of the absolute value of the whole thing,
is just going to become a very, very,
very large number. If r was equal to one,
then the denominator is going to become zero. And we're going to be
dividing by that denominator, and this formula
just breaks down. But where this formula
can be helpful, and where we can get
this to actually give us a sensical result, is when
the absolute value of r is between zero and one. We've already talked
about, we're not even dealing with the
geometric, we're not even talking about a geometric
series if r is equal to zero. So let's think about the case
where the absolute value of r is greater than zero,
and it is less than one. What's going to
happen in that case? Well, the denominator is going
to make sense, right over here. And then up here,
what's going to happen? Well, if you take something
with an absolute value less than one, and you take it
to higher and higher and higher exponents, every time you
multiply it by itself, you're going to get a number
with a smaller absolute value. So this term right over
here, this entire term, is going to go to zero
as n approaches infinity. Imagine if r was 1/2. You're talking about 1/2
to the hundredth power, 1/2 to the thousandth power,
1/2 to the millionth power, 1/2 to the billionth power. That quickly approaches zero. So this goes to zero if
the absolute value of r is less than one. So this, we could argue, would
be equal to a over one minus r. So for example, if I had
the geometric series, if I had the infinite
geometric series-- let's just have a simple one. Let's say that my
first term is one, and then each successive term
I'm going to multiply by 1/3. So it's one plus 1/3
plus 1/3 squared plus 1/3 to the third plus, and I were
to just keep on going forever. This is telling us that that
sum, this infinite sum-- I have an infinite
number of terms here-- this is a pretty
fascinating concept here-- will come out to this. It's going to be my
first term, one, over one minus my common ratio. My common ratio in
this case is 1/3. One minus 1/3, which is the
same thing as one over 2/3, which is equal to 3/2, or you
could view it as one and 1/2. That's a mildly amazing thing.