Explanation Given: The series, 1 r + 1 r 2 + 1 r 3 + ⋯ = 1 r − 1 , | r | > 1 Proof: Consider the left side, 1 r + 1 r 2 + 1 r 3 + ⋯ Since this is an infinite geometric series with, common ratio r 1 = 1 r and a = 1 r . Since, it is given that | r | > 1 , so | 1 r | < 1

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Chapter 9.2, Problem 101E

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To prove: 1r+1r2+1r3+⋯=1r−1, |r|>1.

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Chapter 9.2, Problem 100E

Chapter 9.2, Problem 102E

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Chapter 9 Solutions

EBK CALCULUS: EARLY TRANSCENDENTAL FUNC

Ch. 9.1 - Writing the Terms of a Sequence In exercises 5-10,...Ch. 9.1 - Writing the Terms of a Sequence In Exercises 5-10,...Ch. 9.1 - Writing the Terms of a Sequence In Exercises 5-10,...Ch. 9.1 - Writing the Terms of a Sequence In Exercises 5-10,...Ch. 9.1 - Writing the Terms of a Sequence In exercises 5-10,...Ch. 9.1 - Prob. 6E Ch. 9.1 - Writing the Terms of a Sequence In Exercises 11...Ch. 9.1 - Writing the Terms of a Sequence In Exercises 11...Ch. 9.1 - Prob. 9E Ch. 9.1 - Matching In Exercises 13-16, match the sequence...

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To prove: 1 r + 1 r 2 + 1 r 3 + ⋯ = 1 r − 1 , | r | > 1 .

Solution Summary: The author explains that the geometric series will be convergent. The sum is given by cc

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To prove: 1r+1r2+1r3+⋯=1r−1, |r|>1.

Want to see the full answer?

Chapter 9 Solutions

To prove: 1 r + 1 r 2 + 1 r 3 + ⋯ = 1 r − 1 , | r | &gt; 1 .

Solution Summary: The author explains that the geometric series will be convergent. The sum is given by cc

Videos

To prove: 1r+1r2+1r3+⋯=1r−1, |r|>1.

Want to see the full answer?

Chapter 9 Solutions

To prove: 1 r + 1 r 2 + 1 r 3 + ⋯ = 1 r − 1 , | r | > 1 .