Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove that if \(\sum_{k=1}^{\infty} a_{k}^{2}\) and \(\sum_{k=1}^{\infty} b_{k}^{2}\) both converge then \(\sum_{k=1}^{\infty} a_{k} b_{k}\) converges absolutely. Hint: First show that \(2\left|a_{k} b_{k}\right| \leq a_{k}^{2}+b_{k}^{2}\).

Step-by-Step Solution

Answer

The series \(\sum_{k=1}^{\infty} a_{k} b_{k}\) converges absolutely by comparison.

Step 1: Use the Cauchy-Schwarz Inequality

To prove the convergence of \(\sum_{k=1}^{\infty} a_{k} b_{k}\), we'll first apply the inequality that parallels the Cauchy-Schwarz Inequality: \(2|a_{k} b_{k}| \leq a_{k}^{2} + b_{k}^{2}\). This inequality sets the groundwork for establishing a series of bounds that we will use to prove convergence.

Step 2: Establish the Series Inequality

Knowing the inequality \(2|a_{k} b_{k}| \leq a_{k}^{2} + b_{k}^{2}\), it follows for the whole series: \(\sum_{k=1}^{\infty} 2|a_{k} b_{k}| \leq \sum_{k=1}^{\infty} (a_{k}^{2} + b_{k}^{2})\). Since both \(\sum_{k=1}^{\infty} a_{k}^{2}\) and \(\sum_{k=1}^{\infty} b_{k}^{2}\) converge by hypothesis, their sum, \(\sum_{k=1}^{\infty} (a_{k}^{2} + b_{k}^{2})\), also converges.

Step 3: Conclude Absolute Convergence of Combined Series

Since \(\sum_{k=1}^{\infty} 2|a_{k} b_{k}|\) is less than or equal to a convergent series, it must also converge. Therefore, it follows that \(\sum_{k=1}^{\infty} |a_{k} b_{k}|\) converges (by dividing by 2). A convergent series of absolute values implies that the series \(\sum_{k=1}^{\infty} a_{k} b_{k}\) converges absolutely.

Key Concepts

Cauchy-Schwarz Inequality

The Cauchy-Schwarz Inequality is an important tool in mathematics, especially when dealing with inner product spaces and series convergence. It's a statement about vectors and their inner products, basically giving an upper bound for the absolute value of the dot product of two vectors in terms of the products of their magnitudes. In simpler terms, for any real or complex numbers, the inequality states: \[|a_1 b_1 + a_2 b_2 + ext{{...}} + a_n b_n|^2 \ \leq (a_1^2 + a_2^2 + ext{{...}} + a_n^2)(b_1^2 + b_2^2 + ext{{...}} + b_n^2)\] When working with series, an adapted version of this inequality is used to show that if both sequences \( \{a_k\} \) and \( \{b_k\} \) are squared and summed, and individually converge, then the series formed by their product also satisfies a form of convergence. Specifically, the inequality \(2|a_k b_k| \leq a_k^2 + b_k^2\) helps in setting boundaries that allow us to conclude convergence properties of their product series. This step is crucial as it helps us understand the interactions between the terms of different sequences in a combined series setup.

Absolute Convergence

Absolute convergence is a strong form of convergence for series. We say that a series \( \sum_{k=1}^{\infty} a_k \) converges absolutely if the series of absolute values, \( \sum_{k=1}^{\infty} |a_k| \), converges. This property is desirable because it guarantees convergence independent of term rearrangement, due to the positive nature of absolute values.
In the context of this exercise, we apply the concept of absolute convergence to determine if the series \( \sum_{k=1}^{\infty} a_k b_k \) converges when both \( \sum_{k=1}^{\infty} a_k^2 \) and \( \sum_{k=1}^{\infty} b_k^2 \) are given to converge. Using the inequality \(2|a_k b_k| \leq a_k^2 + b_k^2\), we establish that the sum of a sequence of products of two independent sequences can be bounded by converging series. This powerful result not only shows that the coefficients interact beneficially under multiplication but also ensures that absolute convergence of these products implies the series itself is independently convergent.

Convergent Series

A convergent series is simply a series whose partial sums approach a specific number, known as the sum of the series, as the number of terms increases indefinitely. For a series \( \sum_{k=1}^{\infty} a_k \), if the sequence of partial sums \( S_n = a_1+a_2+\text{...}+a_n \) has a finite limit as \( n \) tends to infinity, we say the series converges.
In this exercise, the convergence of \( \sum_{k=1}^{\infty} a_k b_k \) is demonstrated by leveraging the convergence of \( \sum_{k=1}^{\infty} a_k^2 \) and \( \sum_{k=1}^{\infty} b_k^2 \). Since individual convergence implies that the series formed from the sum \( a_k^2 + b_k^2 \) also converges, it effectively bounds \( \sum_{k=1}^{\infty} 2|a_k b_k| \).
Thus, according to the comparison test – if a larger series converges, a smaller series must also converge – we establish convergence as the series of interest is smaller than a guaranteed convergent one. This method ensures that what can sometimes seem abstract or unapproachable becomes a solid mathematical concept, laying a foundation for understanding complex series interactions.