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Chapter 9: Problem 89

Infinite products Use the ideas of Exercise 88 to evaluate the following infinite products. $$\text { a. } \prod_{k=0}^{\infty} e^{1 / 2^{k}}=e \cdot e^{1 / 2} \cdot e^{1 / 4} \cdot e^{1 / 8} \ldots$$ $$\text { b. } \prod_{k=2}^{\infty}\left(1-\frac{1}{k}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \dots$$

Short Answer

Expert verified
Answer: The value of the first infinite product is $e^2$. Question 2: What is the value of the infinite product $\prod_{k=2}^{\infty}\left(1-\frac{1}{k}\right)$? Answer: The value of the second infinite product is $\frac{1}{3}$.

Step by step solution

01

a. Rewrite the infinite product as a sum of exponents

We can rewrite the expression as an infinite sum of exponents by using the rule \(e^a \cdot e^b = e^{a+b}\): $$\prod_{k=0}^{\infty} e^{1 / 2^{k}} = e^{1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots}$$
02

a. Evaluate the infinite sum

We can now evaluate the infinite sum of exponent: $$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots$$ This is a geometric series with the first term \(a = 1\) and common ratio \(r = \frac{1}{2}\). The sum of an infinite geometric series is given by the formula: $$\frac{a}{1 - r}$$ So the sum is: $$\frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$$
03

a. Final answer for the first infinite product

We can now plug the sum back into the expression for the infinite product: $$e^{1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots} = e^{2}$$ So the value of the first infinite product is \(\boxed{e^2}\).
04

b. Simplify the telescoping product

We notice that the product is telescoping, which means terms will cancel each other out. Let's rewrite the expression to see which terms cancel each other out: $$\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \dots = \frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \dots$$
05

b. Cancel the terms and find the final answer

We see that the terms in the product cancel each other: $$\frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \dots = \frac{1}{3} \cdot \left(\frac{1}{2}\cdot\frac{2}{4}\right) \cdot\left(\frac{2}{3}\cdot\frac{3}{5}\right)\dots$$ As \(n \to \infty\), we are left with the first term which is \(\boxed{\frac{1}{3}}\). So the value of the second infinite product is \(\boxed{\frac{1}{3}}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometric Series
When dealing with a geometric series, we encounter sequences where each term is a constant multiple of the previous one. This fixed number is known as the **common ratio**. For example, in the series mentioned in the exercise: \[1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots\] The first term is 1, and the common ratio is \(r = \frac{1}{2}\). When you repeatedly multiply by the common ratio, you can see how each term progresses.
A vital property of geometric series is that if the absolute value of the common ratio is less than 1, the series will converge to a sum. The formula for an infinite geometric series is:
  • Sum = \(\frac{a}{1-r}\) where \(a\) is the first term and \(r\) is the common ratio.
In our exercise, substituting \(a = 1\) and \(r = \frac{1}{2}\), the series sums to 2, which allows us to find \(e^2\) for the infinite product.
Telescoping Series
A telescoping series is a type of series where successive terms cancel one another out. This makes it easier to evaluate because many terms fade away, leaving behind only a few to consider. It's like looking through a telescope, focusing on one small piece at the end of a long tube.
In the exercise, the product:\[\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots\]can be rearranged to show how terms cancel:\[\frac{1}{2} \cancel{\cdot \frac{2}{3} \cdot \frac{3}{4}} \cdot \cancel{\frac{3}{3} \cdot \frac{4}{4}} \cdots\]Many intermediate terms disappear when you break down the fractions, leaving mostly the first and last terms. As \(n\) goes to infinity, these series often resolve into much simpler forms.
Exponentiation
Exponentiation refers to the mathematical operation involving numbers raised to a power. It's like a repeated multiplication of a number by itself. In simpler terms, if \(a^n\) translates to multiplying \(a\) by itself \(n\) times.
In the context of this exercise, exponentiation has been used to handle products of expressions involving exponents. Look at this progression:\[e^{1/2^0} \cdot e^{1/2^1} \cdot e^{1/2^2} \cdot \ldots = e^{1 + 1/2 + 1/4 + \ldots}\]By understanding the sum of an infinite geometric series through exponentiation, this infinite product simplifies into a compact form, i.e., \(e^2\).
  • Remember: Product of exponents with the same base means addition of the powers: \(e^a \cdot e^b = e^{a+b}\).
These concepts of repeated multiplication and addition through exponentiation make complex infinite products more manageable.

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Most popular questions from this chapter

a. Consider the number 0.555555...., which can be viewed as the series \(5 \sum_{k=1}^{\infty} 10^{-k} .\) Evaluate the geometric series to obtain a rational value of \(0.555555 \ldots\) b. Consider the number \(0.54545454 \ldots,\) which can be represented by the series \(54 \sum_{k=1}^{\infty} 10^{-2 k} .\) Evaluate the geometric series to obtain a rational value of the number. c. Now generalize parts (a) and (b). Suppose you are given a number with a decimal expansion that repeats in cycles of length \(p,\) say, \(n_{1}, n_{2} \ldots \ldots, n_{p},\) where \(n_{1}, \ldots, n_{p}\) are integers between 0 and \(9 .\) Explain how to use geometric series to obtain a rational form of the number. d. Try the method of part (c) on the number \(0.123456789123456789 \ldots\) e. Prove that \(0 . \overline{9}=1\)

Consider the expression \(\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}},\) where the process continues indefinitely. a. Show that this expression can be built in steps using. the recurrence relation \(a_{0}=1, a_{n+1}=\sqrt{1+a_{n}},\) for \(n=0,1,2,3, \ldots .\) Explain why the value of the expression can be interpreted as \(\lim _{n \rightarrow \infty} a_{n}\). b. Evaluate the first five terms of the sequence \(\left\\{a_{n}\right\\}\). c. Estimate the limit of the sequence. Compare your estimate with \((1+\sqrt{5}) / 2,\) a number known as the golden mean. d. Assuming the limit exists, use the method of Example 5 to determine the limit exactly. e. Repeat the preceding analysis for the expression \(\sqrt{p+\sqrt{p+\sqrt{p+\sqrt{p+\cdots}}}}\) where \(p > 0 .\) Make a table showing the approximate value of this expression for various values of \(p .\) Does the expression seem to have a limit for all positive values of \(p ?\)

Evaluate the limit of the following sequences. $$a_{n}=\cos \left(0.99^{n}\right)+\frac{7^{n}+9^{n}}{63^{n}}$$

After many nights of observation, you notice that if you oversleep one night you tend to undersleep the following night, and vice versa. This pattern of compensation is described by the relationship $$x_{n+1}=\frac{1}{2}\left(x_{n}+x_{n-1}\right), \quad \text { for } n=1,2,3, \ldots$$ where \(x_{n}\) is the number of hours of sleep you get on the \(n\) th night and \(x_{0}=7\) and \(x_{1}=6\) are the number of hours of sleep on the first two nights, respectively. a. Write out the first six terms of the sequence \(\left\\{x_{n}\right\\}\) and confirm that the terms alternately increase and decrease. b. Show that the explicit formula $$x_{n}=\frac{19}{3}+\frac{2}{3}\left(-\frac{1}{2}\right)^{n}, \text { for } n \geq 0,$$ generates the terms of the sequence in part (a). c. What is the limit of the sequence?

A glimpse ahead to power series Use the Ratio Test to determine the values of \(x \geq 0\) for which each series converges. $$\sum_{k=1}^{\infty} \frac{x^{k}}{2^{k}}$$
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