91Ó°ÊÓ

Infinite products An infinite product \(P=a_{1} a_{2} a_{3} \ldots,\) which is denoted \(\prod_{k=1}^{\infty} a_{k}\) is the limit of the sequence of partial products \(\left\\{a_{1}, a_{1} a_{2}, a_{1} a_{2} a_{3}, \dots\right\\}\) a. Show that the infinite product converges (which means its sequence of partial products converges) provided the series \(\sum_{k=1}^{\infty} \ln a_{k}\) converges. b. Consider the infinite product $$P=\prod_{k=2}^{\infty}\left(1-\frac{1}{k^{2}}\right)=\frac{3}{4} \cdot \frac{8}{9} \cdot \frac{15}{16} \cdot \frac{24}{25} \cdots$$ Write out the first few terms of the sequence of partial products, $$P_{n}=\prod_{k=2}^{n}\left(1-\frac{1}{k^{2}}\right)$$ (for example, \(P_{2}=\frac{3}{4}, P_{3}=\frac{2}{3}\) ). Write out enough terms to determine the value of the product, which is \(\lim _{n \rightarrow \infty} P_{n}\). c. Use the results of parts (a) and (b) to evaluate the series $$\sum_{k=2}^{\infty} \ln \left(1-\frac{1}{k^{2}}\right)$$

Step by step solution

01

Prove the convergence of the infinite product

To show that the infinite product \(P=a_{1} a_{2} a_{3} \ldots\) converges provided the series \(\sum_{k=1}^{\infty} \ln a_{k}\) converges, consider the logarithm of the partial products of the infinite product: $$ \ln (a_{1} a_{2} a_{3} \ldots a_{n}) = \ln a_{1} + \ln a_{2} + \ln a_{3} + \ldots + \ln a_{n} $$ By the properties of logarithms, the sum of individual logarithms of each term is equal to the logarithm of their product. Now, if we take the limit as \(n \rightarrow \infty\), we have the series: $$ \lim_{n \rightarrow \infty} \ln (a_{1} a_{2} a_{3} \ldots a_{n}) = \sum_{k=1}^{\infty} \ln a_{k} $$ If the series converges, i.e., if \(\sum_{k=1}^{\infty} \ln a_{k} = L\) for some finite value of L, we can write: $$ \lim_{n \rightarrow \infty} \ln (a_{1} a_{2} a_{3} \ldots a_{n}) = L $$ Applying the exponential function: $$ \lim_{n \rightarrow \infty} (a_{1} a_{2} a_{3} \ldots a_{n}) = e^{L} $$ Since both sides are finite, this means that the infinite product converges.

02

Determine the value of the given infinite product

Consider the given infinite product: $$ P=\prod_{k=2}^{\infty}\left(1-\frac{1}{k^{2}}\right) $$ We need to find the first few terms of its partial products: \begin{align*} P_2 &= \frac{3}{4}\\ P_3 &= \frac{3}{4} \cdot \frac{8}{9} = \frac{2}{3}\\ P_4 &= \frac{3}{4} \cdot \frac{8}{9} \cdot \frac{15}{16} = \frac{1}{2} \end{align*} Observe that the pattern is \(P_n = \frac{n-1}{n}\), so the value of the product is the limit as \(n\to\infty\) of the sequence \(P_n\): $$ \lim_{n \rightarrow \infty} P_n = \lim_{n \rightarrow \infty} \frac{n-1}{n} = 1 $$

03

Evaluate the given series

Using the result from Step 1, we can evaluate the series: $$ \sum_{k=2}^{\infty} \ln \left(1-\frac{1}{k^{2}}\right) $$ By using the convergence relation we obtained in Step 1, we can write: $$ \ln \left(\prod_{k=2}^{\infty}\left(1-\frac{1}{k^{2}}\right)\right) = \sum_{k=2}^{\infty} \ln \left(1-\frac{1}{k^{2}}\right) $$ We already found the limit of the product in Step 2, which is 1: $$ \ln(1) = \sum_{k=2}^{\infty} \ln \left(1-\frac{1}{k^{2}}\right) $$ Since \(\ln(1) = 0\), our final answer is: $$ \sum_{k=2}^{\infty} \ln \left(1-\frac{1}{k^{2}}\right) = 0 $$

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Key Concepts

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

Convergent Series

In mathematics, a convergent series is a series whose partial sums tend towards a specific limit as the number of terms increases. This means if you keep adding up the terms in the series, they will get closer and closer to a particular number rather than diverging to infinity or oscillating indefinitely.

To understand convergence, think of it as the process of getting closer to a target. If you have a series \( a_1 + a_2 + a_3 + \ldots \), and after adding increasingly many terms, the sum approaches a fixed number, the series is said to converge.

One essential aspect of determining convergence for series is to check the behavior of the terms. If the terms become very small as you proceed in the series, there could be a chance that the series converges. However, this condition alone doesn't guarantee convergence; the sum of the terms must approach a finite limit.

Logarithmic Functions

Logarithmic functions are inverses of exponential functions and are fundamental in mathematics, especially when dealing with series and products like we do with infinite products. The logarithm of a number is the exponent to which the base, usually 10 or e, must be raised to produce that number.

For example, \(lge{e}{y} = x \) means \( e^x = y \). Logarithms are incredibly useful because they transform products into sums, which are easier to manage. In the context of infinite products and series, using logarithms helps us to analyze convergence in a more manageable form.

For infinite products, the convergence can be shown if the sum of the logarithms of the terms converges. This property is derived from the fact that:\[\ln(a_1) + \ln(a_2) + \ln(a_3) + \ldots + \ln(a_n) = \ln(a_1 \times a_2 \times a_3 \times \ldots \times a_n)\]Using this property simplifies complex multiplication problems into addition problems, which are generally simpler to solve.

Partial Products

Partial products are intermediate steps used when evaluating infinite products. They are essentially finite products of an infinite sequence. If the partial products converge as more terms are included, the entire infinite product is said to converge.

For example, when considering an infinite product like \(\prod_{k=2}^{\infty}\left(1-\frac{1}{k^2}\right)\), the partial products would look like:

• \(P_2 = \frac{3}{4}\)
• \(P_3 = \frac{3}{4} \times \frac{8}{9} = \frac{2}{3}\)
• \(P_4 = \frac{3}{4} \times \frac{8}{9} \times \frac{15}{16} = \frac{1}{2}\)

Notice how the partial product sequence \( P_n = \frac{n-1}{n} \) gets closer to 1 as \( n \to \infty \). This is how we determine the convergence of the infinite product by observing the limit of its partial products.

Exponential Function

Exponential functions are vital in understanding the convergence of infinite products, as they are the inverse operations to logarithmic functions. The exponential function raises a constant base to the power of the given number, usually written as \( e^x \), where \( e \) is approximately equal to 2.718.

In mathematics, exponentials are used for solving problems involving growth and decay but also play a crucial role in mathematical transformations. They help revert logarithmic results back to multiplicative form.

In the context of infinite products, after showing using logarithms that a series converges to a particular number \( L \), the exponential function allows us to find the original product by calculating \( e^L \). This step is critical because applying \( \exp \) to the logarithm of a product allows us to retrieve the product's actual limit when its series of logarithms is convergent.