where did you get the -ar^(n+1), that's not in first video

The video is here: https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-factor/x2ec2f6f830c9fb89:geo-series/v/deriving-formula-for-sum-of-finite-geometric-series

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)

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".

Why is the second proof more formal than the first? Is it the approach? Starting from a finite sum and taking an infinite limit?

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.

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.

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.

Main content

Integral Calculus

Course: Integral Calculus > Unit 5

Lesson 2: Infinite geometric series

Proof of infinite geometric series formula

AP.CALC:

LIM‑7 (EU)

LIM‑7.A (LO)

LIM‑7.A.3 (EK)

LIM‑7.A.4 (EK)

Google Classroom

Say we have an infinite geometric series whose first term is a and common ratio is r. If r is between minus, 1 and 1 (i.e. vertical bar, r, vertical bar, is less than, 1), then the series converges into the following finite value:

limit, start subscript, n, \to, infinity, end subscript, sum, start subscript, i, equals, 0, end subscript, start superscript, n, end superscript, a, dot, r, start superscript, i, end superscript, equals, start fraction, a, divided by, 1, minus, r, end fraction

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.

Khan Academy video wrapper

Infinite geometric series formula intuitionSee video transcript

Now we can prove the formula more formally.

Khan Academy video wrapper

Proof of infinite geometric series as a limitSee video transcript

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Soldier King
Posted 4 years ago. Direct link to Soldier King's post “where did you get the -ar...”
where did you get the -ar^(n+1), that's not in first video
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(19 votes)
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- Adil
  Posted 2 years ago. Direct link to Adil's post “The video is here: https:...”
  The video is here: https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-factor/x2ec2f6f830c9fb89:geo-series/v/deriving-formula-for-sum-of-finite-geometric-series
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Sepehr Shariat panahi
Posted 5 years ago. Direct link to Sepehr Shariat panahi's post “why |r| must be lower tha...”
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.
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(3 votes)
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- kubleeka
  Posted 5 years ago. Direct link to kubleeka's post “If |r|≥1, then the size o...”
  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.
  Comment on kubleeka's post “If |r|≥1, then the size o...”
  (11 votes)
essejnosnevets
Posted 5 years ago. Direct link to essejnosnevets's post “Why is the second proof m...”
Why is the second proof more formal than the first? Is it the approach? Starting from a finite sum and taking an infinite limit?
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(4 votes)
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- kubleeka
  Posted 5 years ago. Direct link to kubleeka's post “The first proof was less ...”
  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.
  Comment on kubleeka's post “The first proof was less ...”
  (6 votes)
Michele Franzoni
Posted 3 years ago. Direct link to Michele Franzoni's post “Isn't the formula for a g...”
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)
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(5 votes)
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- Elliott Walker
  Posted 2 years ago. Direct link to Elliott Walker's post “I know this is 6 months l...”
  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".
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  (3 votes)
naiyar2006
Posted 2 months ago. Direct link to naiyar2006's post “Is my proof for the formu...”
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)
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(3 votes)
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- Venkata
  Posted 2 months ago. Direct link to Venkata's post “Yep! Seems good to me!”
  Yep! Seems good to me!
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  (3 votes)
Redmond Riddell
Posted 3 years ago. Direct link to Redmond Riddell's post “In the first video, if ev...”
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..
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(2 votes)
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