Question

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

Short answer

Question: Evaluate the following infinite products: (a) $$\prod_{k=0}^{\infty} e^{1 / 2^{k}}$$ and (b) $$\prod_{k=2}^{\infty}\left(1-\frac{1}{k}\right)$$. Answer: (a) The evaluation of the infinite product is $$e^{2}$$. (b) The evaluation of the infinite product is $$1$$.

Step-by-step solution

Simplify expression (a)

Observe that the terms in the product of expression (a) involve the common base of $$e$$ raised to an exponent that is a power of 2 (starting from $$k=0$$). We can use the properties of exponents to merge the terms into a single power of $$e$$. Specifically, the rule here is $$e^{a}e^{b}=e^{(a+b)}$$. The expression becomes: $$\prod_{k=0}^{\infty} e^{1 / 2^{k}}=e^{1/2^{0}} \cdot e^{1 / 2^{1}} \cdot e^{1 / 2^{2}} \cdot e^{1 / 2^{3}}\cdots$$ Then apply the exponent rule and merge the terms into a single power of $$e$$: $$e^{(1/2^{0}+1/2^{1}+1/2^{2}+1/2^{3}+\cdots)}$$

Evaluate the sum of the exponent of (a)

Now, we need to evaluate the sum of the exponent: $$1/2^{0}+1/2^{1}+1/2^{2}+1/2^{3}+\cdots$$ This is a geometric series with the first term $$a_1=1$$ and the common ratio $$r=1/2$$. To find the sum of this infinite geometric series, we use the formula: $$\text{sum} = \frac{a_1}{1 - r} = \frac{1}{1 - (1/2)}$$ Calculate the sum: $$\text{sum} = \frac{1}{1/2}=2$$

Substitute the sum back into the expression of (a)

Now that we have determined the sum of the exponent in the simplified expression, we can substitute the sum back into the expression: $$e^{(1/2^{0}+1/2^{1}+1/2^{2}+1/2^{3}+\cdots)}=e^{2}$$ The evaluation of the infinite product of expression (a) is $$e^{2}$$.

Simplify expression (b)

Expression (b) is a telescoping product, which cancels out adjacent terms when expanded: $$\prod_{k=2}^{\infty}\left(1-\frac{1}{k}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdots$$ Each term except the first denominator and the last numerator cancels out.

Find the limit of the product of expression (b)

We are left with the simplified infinite product: $$\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdot\frac{4}{6}\cdots$$ We can rewrite each fraction as: $$\frac{k-1}{k+1}$$ Now, we need to find the limit of this product as $$k$$ approaches infinity: $$\lim_{k\to\infty} \frac{k-1}{k+1}$$ As $$k$$ becomes very large, the fractions tend towards $$1$$, and their product also tends towards $$1$$. Hence, the evaluation of the infinite product of expression (b) is $$1$$.

Key concepts

Geometric Series

A geometric series is formed when each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. An infinite geometric series can be summed using a special formula, provided that the absolute value of the common ratio is less than 1.

For instance, the series \(\sum_{k=0}^{\infty} r^k\) is geometric, and its sum is \(\frac{a_1}{1-r}\) where \(a_1\) is the first term and \(r\) is the common ratio. This formula is derived from the fact that each term (except the first) in a geometric series can be expressed as the previous term multiplied by the common ratio. In our example exercise, the sum of the exponents in expression (a) forms an infinite geometric series with a common ratio of \(\frac{1}{2}\), leading to the sum being \(2\).

To help students grasp this concept, it is essential to clarify the conditions for the convergence of an infinite geometric series and to explain the summation process with real examples.

Properties of Exponents

The properties of exponents are rules that describe how to handle powers of numbers or expressions when they are multiplied, divided, or raised to another power. One of the fundamental rules used in our exercise solution is that when we multiply expressions with the same base, we can add their exponents.

In mathematical terms, \(e^a \cdot e^b = e^{a+b}\). This rule makes it possible to merge multiple exponential terms into a single exponent, simplifying the expression significantly. For example, in the exercise, we combined the exponents of the terms \(e^{1/2^k}\) into a single exponent. Understanding and applying these exponent rules allows students to manipulate and simplify complex expressions, as seen in part (a) of our exercise.

Telescoping Product

A telescoping product, much like a telescoping series, refers to an expression where after expansion, many terms cancel out, leaving behind a simplified form. This happens when consecutive factors in a product have reciprocal elements that diminish upon multiplication. In the context of the provided exercise, the product \(\prod_{k=2}^{\infty}(1-\frac{1}{k})\) can be expanded to show that each term \(\frac{k}{k+1}\) in the product will cancel the \(k+1\) term from the following term's \(k-1\) in the numerator.

Understanding a telescoping product is best demonstrated by explicitly writing out the first few terms of the product and observing how the cancellation occurs. In the end, we are typically left with just a few terms or even a single term, which greatly simplifies the evaluation of the product, as seen in part (b) of our exercise.

Limits

The concept of limits is foundational in calculus and pertains to the behavior of a function as the input approaches a certain value. When analyzing infinite products or series, finding the limit helps determine the value the expression approaches as the number of terms grows indefinitely. In the second part of our exercise \(b\), we find \(\lim_{k\to\infty} (\frac{k-1}{k+1})\) to evaluate the infinite product. As \(k\) increases, the value of \(\frac{k-1}{k+1}\) approaches 1, because the terms \(k-1\) and \(k+1\) become negligible compared to the ever-increasing \(k\).

Teaching the concept of limits usually involves discussing the idea of approaching rather than arriving at a value, and how this concept applies to sequences, series, and functions. Applying limits to infinite products or series like in our exercise involves identifying the dominant terms as the sequence or series extends towards infinity.