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Proof of infinite geometric series as a limit

Sal applies limits to the formula for the sum of a finite geometric series to get the sum of an infinite geometric series. Created by Sal Khan.

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• What would happen if the absolute value of r wasn't between 0 and 1? Would the sum just be infinite?
• 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".
• So this equation only works when 'r' is between zero and one?
• Actually |r|<1 i.e -1<r<1. This the required condition for the equation to work. Hope it helped
• What does Zeno's paradox has to do with this?
• 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.
• How can you find a common 'r' if the first number in the series is zero?
• 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.
• 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.
• , 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?
• 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 around .
• I thought the formula is a(1-r^n)/1-r
why did it expands to n+1 ? (a-ar^n+1)/1-r
?
• How do you find the ratio if it isn't given?
• 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.