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### Course: 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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# Proof of infinite geometric series formula

Say we have an infinite geometric series whose first term is $a$ and common ratio is $r$ . If $r$ is between $-1$ and $1$ (i.e. $|r|<1$ ), then the series converges into the following finite value:

The AP Calculus course doesn't require knowing the proof of this fact, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn.

## First, let's get some intuition for why this is true. This isn't a formal proof but it's quite insightful.

## Now we can prove the formula more formally.

## Want to join the conversation?

- where did you get the -ar^(n+1), that's not in first video(22 votes)
- why |r| must be lower than 1?

mathematically its correct that we reach to some specific value for S sub infinity value. But it is negative.

I know that for r=1 this formula cant help us.(3 votes)- If |r|≥1, then the size of the terms doesn't decrease. So for any value you might think this converges to, it eventually exceeds it.(14 votes)

- Why is the second proof more formal than the first? Is it the approach? Starting from a finite sum and taking an infinite limit?(4 votes)
- The first proof was less formal because we assumed that the sum converged. That's a necessary thing to assume/prove if we're going to treat S like any other real number that can be moved around the equation.

The second proof was more rigorous because we didn't assume |r|<1, and explored what would happen to the sum if r took on different values.(7 votes)

- Isn't the formula for a geometric series equal to

a(1-r^n)/(1-r) ?

This is the result Sal found in the related video (check it out here https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-factor/x2ec2f6f830c9fb89:geo-series/v/deriving-formula-for-sum-of-finite-geometric-series)(6 votes)- I know this is 6 months late, but whatever- That's the sum of a finite geometric series. This formula is for the sum of an INFINITE geometric series, which returns the output given what is essentially an infinite "n".(5 votes)

- Is my proof for the formula of a finite geometric series correct:

(1) Sn = a + ar + ar^2 + ... + ar^n-1

Sn - a = ar + ar^2 + ... + ar^n-1

(2) (Sn - a)/r = a + ar + ... + ar^n-2

(1-2) Sn - (Sn-a)/r = ar^n-1

Now solve for Sn:

(rSn - Sn + a)/r = ar^n-1

Sn(r-1)/r + a/r = ar^n-1

Sn(r-1) + a = (ar^n-1)r = ar^n

Sn = a(r^n - 1)/(r-1)(4 votes)- Yep! Seems good to me!(3 votes)

- In the first video, if every a in the first equation gets cancelled out by an a in the second equation (assuming this is true for infinity, since both equations go to infinity) except for the initial ar^0, how does this formula not work for r>1?

I realise the denominator becomes a negative, and this doesn't necessarily make sense.. but what is the reason r is limited in the values it can take on..? it isn't implicit in the derivation.. it just is, because Sal says so! is this because we are dealing with with infinity?

I understand when r < 1, eventually our sum will converge to a number, and this makes sense. but what about this derivation limits the scope of r? Intuitively, I get that a series will NOT converge if r > 1... but surely we need an implicit reason that this works for some numbers, and not others? The way it is written, is that the second equations huge number should cancel out the same huge number in the first equation, and we will be left with an a.. but this does not make sense, since the series shouldn't converge..(2 votes) - in the second video above, it starts with the sum of a finite geometric series=(a-ar^(n+1))/1-r.

but in another finite geometric series formula video (https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-factor/x2ec2f6f830c9fb89:geo-series/v/deriving-formula-for-sum-of-finite-geometric-series), the formula is (a-ar^n)/1-r.

Why the difference?(1 vote)- Disclaimer: I'm not 100% sure.

In this summation, it starts from 0 to n**inclusively**. It means there are n + 1 terms. In the other video, there are n terms.

It's still the same formula.(3 votes)

- this is one of the few topics where the wikipedia article is legible(1 vote)