Explanation Given information: A perpetuity is promises to pay P dollars at the end of each month. (The first payment will be received in 1 month.) If the interest rate per month is r , the present value of P dollars in k month is P ( 1 + r ) − k Explanation: The perpetuity pay P dollars at the beginning of each month at the interest rate of r % compounded monthly, so the present value of P dollars in first month is P ( 1 + r ) − 1 . For the second month, the present value is P ( 1 + r ) − 1 ( 1 + r ) − 1 = P ( 1 + r ) − 2 , and so on. The present value of P dollars in k months is an infinite geometric series. P ( 1 + r ) 0 + P ( 1 + r ) − 1 + P ( 1 + r ) − 2 + P ( 1 + r ) − 3 + . . . + P ( 1 + r ) − k This series, in summation notation is written as: ∑ k = 1 ∞ P ( 1 + r ) − k Thus, the capital value of the perpetuity as an infinite series is the sum of the present value of P dollars in infinite months. The capital value of the perpetuity as an infinite series is ∑ k = 1 ∞ P ( 1 + r ) − k ∑ k = 1 ∞ P ( 1 + r ) − k = ∑ k = 0 ∞ P ( 1 + r ) − k − P ∑ k = 0 ∞ P ( 1 + r ) − k is a geometric series with initial term a = P and the common ratio | r | = | 1 ( 1 + r ) | = 1 ( 1 + r ) < 1

15th Edition

ISBN: 9780137590896

Author: Larry J. Goldstein; David C. Lay; David I. Schneider; Nakhle H. Asmar; William Edward Tavernetti

Publisher: Pearson Education (US)

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Question

Chapter 11.3, Problem 26E

To determine

A simple formula for the capital value of the perpetuity, where a perpetuity that

promises to pay $P$ dollars at the end of each month. (The first payment will be received in $1$

month.) If the interest rate per month is $r$ , the present value of $P$ dollars in $k$ month is

$P(1+r)−k$ .

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Chapter 11.3, Problem 25E

Chapter 11.3, Problem 27E

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Chapter 11 Solutions

Calculus & Its Applications

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Ch. 11.1 - Prob. 9E Ch. 11.1 - Prob. 10E Ch. 11.1 - Prob. 11E Ch. 11.1 - Prob. 12E Ch. 11.1 - Prob. 13E Ch. 11.1 - Prob. 14E Ch. 11.1 - Prob. 15E Ch. 11.1 - Prob. 16E Ch. 11.1 - Prob. 17E Ch. 11.1 - Prob. 18E Ch. 11.1 - Determine the third and fourthTaylor polynomial...Ch. 11.1 - Prob. 20E Ch. 11.1 - Prob. 21E Ch. 11.1 - Prob. 22E Ch. 11.1 - Prob. 23E Ch. 11.1 - Prob. 24E Ch. 11.1 - Prob. 25E Ch. 11.1 - Prob. 26E Ch. 11.1 - Prob. 27E Ch. 11.1 - Prob. 28E Ch. 11.1 - Prob. 29E Ch. 11.1 - Prob. 30E Ch. 11.1 - Graph the function Y1=11x and its fourth Taylor...Ch. 11.1 - Prob. 32E Ch. 11.1 - Prob. 33E Ch. 11.1 - Prob. 34E Ch. 11.2 - Prob. 1CYU Ch. 11.2 - Prob. 2CYU Ch. 11.2 - In Exercises 18, use three repetitions of the...Ch. 11.2 - In Exercises 18, use three repetitions of the...Ch. 11.2 - Prob. 3E Ch. 11.2 - Prob. 4E Ch. 11.2 - In Exercises 18, use three repetitions of the...Ch. 11.2 - Prob. 6E Ch. 11.2 - Prob. 7E Ch. 11.2 - Prob. 8E Ch. 11.2 - Sketch the graph of y=x3+2x+2, and use the...Ch. 11.2 - 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A simple formula for the capital value of the perpetuity, where a perpetuity that promises to pay P dollars at the end of each month. (The first payment will be received in 1 month.) If the interest rate per month is r , the present value of P dollars in k month is P ( 1 + r ) − k .

Solution Summary: The author explains how a perpetuity promises to pay P dollars at the end of each month.

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A simple formula for the capital value of the perpetuity, where a perpetuity that

promises to pay $P$ dollars at the end of each month. (The first payment will be received in $1$

month.) If the interest rate per month is $r$ , the present value of $P$ dollars in $k$ month is

$P(1+r)−k$ .

Want to see the full answer?

Chapter 11 Solutions