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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

The Riemann zeta function is the subject of extensive research and is associated with several renowned unsolved problems. It is defined by \(\zeta(x)=\sum_{k=1}^{\infty} \frac{1}{k^{x}}\). When \(x\) is a real number, the zeta function becomes a \(p\) -series. For even positive integers \(p,\) the value of \(\zeta(p)\) is known exactly. For example, $$ \sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}, \quad \sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90}, \quad \text { and } \quad \sum_{k=1}^{\infty} \frac{1}{k^{6}}=\frac{\pi^{6}}{945}, \ldots $$ Use estimation techniques to approximate \(\zeta(3)\) and \(\zeta(5)\) (whose values are not known exactly) with a remainder less than \(10^{-3}\).

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the value of the limit or determine that the limit does not exist. $$a_{n+1}=\sqrt{2+a_{n}} ; a_{0}=1, n=0,1,2, \dots$$

Consider the formulas for the following sequences. Using a calculator, make a table with at least 10 terms and determine a plausible value for the limit of the sequence or state that it does not exist. $$a_{n}=\frac{(n-1)^{2}}{\left(n^{2}-1\right)} ; n=2,3,4, \dots$$

Consider the following convergent series. a. Find an upper bound for the remainder in terms of \(n\) b. Find how many terms are needed to ensure that the remainder is less than \(10^{-3}\) c. Find lower and upper bounds $$\left(L_{n} \text { and } U_{n}\right.$$ respectively) on the exact value of the series. d. Find an interval in which the value of the series must lie if you approximate it using ten terms of the series. $$\sum_{k=1}^{\infty} \frac{1}{k^{8}}$$

Suppose a ball is thrown upward to a height of \(h_{0}\) meters. Each time the ball bounces, it rebounds to a fraction r of its previous height. Let \(h_{n}\) be the height after the nth bounce and let \(S_{n}\) be the total distance the ball has traveled at the moment of the nth bounce. a. Find the first four terms of the sequence \(\left\\{S_{n}\right\\}\) b. Make a table of 20 terms of the sequence \(\left\\{S_{n}\right\\}\) and determine a plausible value for the limit of \(\left\\{S_{n}\right\\}.\) $$h_{0}=20, r=0.5$$
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