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How do you find the sum of an infinite geometric series? - Answers
An infinite geometric series can be summed only if the common ratio has an absolute value less than 1. Suppose the sum to n terms is S(n). That is, S(n) = a + ar + ar2 + ... + arn-1 Multipying through by the common ratio, r, gives r*S(n) = ar + ar2 + ar3 + ... + arn Subtracting the second equation from the first, S(n) - r*S(n) = a - arn (1 - r)*S(n) = a*(1 - rn) Dividing by (1 - r), S(n) = (1 - rn)/(1 - r) Now, since |r| < 1, rn tends to 0 as n tends to infinity and so S(n) tends to 1/(1 - r) or, the infinite sum is 1/(1 - r)
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How do you find the sum of an infinite geometric series? - Answers
An infinite geometric series can be summed only if the common ratio has an absolute value less than 1. Suppose the sum to n terms is S(n). That is, S(n) = a + ar + ar2 + ... + arn-1 Multipying through by the common ratio, r, gives r*S(n) = ar + ar2 + ar3 + ... + arn Subtracting the second equation from the first, S(n) - r*S(n) = a - arn (1 - r)*S(n) = a*(1 - rn) Dividing by (1 - r), S(n) = (1 - rn)/(1 - r) Now, since |r| < 1, rn tends to 0 as n tends to infinity and so S(n) tends to 1/(1 - r) or, the infinite sum is 1/(1 - r)
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How do you find the sum of an infinite geometric series? - Answers
An infinite geometric series can be summed only if the common ratio has an absolute value less than 1. Suppose the sum to n terms is S(n). That is, S(n) = a + ar + ar2 + ... + arn-1 Multipying through by the common ratio, r, gives r*S(n) = ar + ar2 + ar3 + ... + arn Subtracting the second equation from the first, S(n) - r*S(n) = a - arn (1 - r)*S(n) = a*(1 - rn) Dividing by (1 - r), S(n) = (1 - rn)/(1 - r) Now, since |r| < 1, rn tends to 0 as n tends to infinity and so S(n) tends to 1/(1 - r) or, the infinite sum is 1/(1 - r)
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- og:descriptionAn infinite geometric series can be summed only if the common ratio has an absolute value less than 1. Suppose the sum to n terms is S(n). That is, S(n) = a + ar + ar2 + ... + arn-1 Multipying through by the common ratio, r, gives r*S(n) = ar + ar2 + ar3 + ... + arn Subtracting the second equation from the first, S(n) - r*S(n) = a - arn (1 - r)*S(n) = a*(1 - rn) Dividing by (1 - r), S(n) = (1 - rn)/(1 - r) Now, since |r| < 1, rn tends to 0 as n tends to infinity and so S(n) tends to 1/(1 - r) or, the infinite sum is 1/(1 - r)
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