Question

Prove: If \(0

Short answer

Question: Prove that if a sequence of positive real numbers \(a_n\) is strictly decreasing and the series \(\sum a_n^k\) converges for some positive integer \(k\), then the series \(\sum (-1)^n \sin(a_nx)\) converges uniformly on any finite interval. Solution: We apply the Weierstrass M-test, which requires us to find a sequence of positive constants \(M_n\) such that \(\lvert (-1)^n \sin(a_nx) \rvert \leq M_n\) and \(\sum M_n < \infty\). We first show that the series \(\sum a_n^{k+1}\) also converges using the comparison test. Then, we choose \(M_n = a_n^{k+1}\) and verify that the conditions of the Weierstrass M-test hold. Thus, the series \(\sum (-1)^n \sin(a_nx)\) converges uniformly on any finite interval.

Step-by-step solution

Prove that the series \(\sum a_n^{k+1}\) converges

Let's consider the series \(\sum a_n^{k+1}\). Observe that since \(0 a_{n+1}^{k+1} > 0\). We can apply the comparison test, which states that if we have a convergent series \(\sum b_n\) and a sequence \(0 \leq a_n \leq b_n\) for all \(n\), then the series \(\sum a_n\) converges as well. Here, we let \(b_n = a_n^k\). Therefore, \(0 \leq a_n^{k+1} \leq a_n^k\). Since \(\sum a_n^{k}<\infty\), the comparison test implies that \(\sum a_n^{k+1}<\infty\) as well.

Find the sequence \(M_n\) to apply the Weierstrass M-test

Now we want to apply the Weierstrass M-test to the series \(\sum (-1)^n \sin(a_nx)\). We first need to find a sequence of positive constants \(M_n\) such that \(\lvert (-1)^n \sin(a_nx) \rvert \leq M_n\). It is clear that \(\lvert \sin(a_nx) \rvert \leq 1\) for all \(n\). Thus, we can choose \(M_n = a_n^{k+1}\), so that \(\lvert (-1)^n \sin(a_nx) \rvert \leq a_n^{k+1}\) for all \(n\).

Verify the Weierstrass M-test conditions and conclude uniform convergence

We have found the sequence \(M_n = a_n^{k+1}\) and shown that \(\lvert (-1)^n \sin(a_nx) \rvert \leq M_n\). We have also proved that the series \(\sum a_n^{k+1}\) converges. Therefore, by the Weierstrass M-test, the series \(\sum (-1)^n \sin(a_nx)\) converges uniformly on any finite interval.

Key concepts

Uniform Convergence

Uniform convergence is a powerful concept in real analysis that provides a way to understand the behavior of series of functions over intervals. Unlike pointwise convergence, where each point approaches the limit individually, uniform convergence ensures that a series of functions converges to a single function uniformly over the entire domain.

This distinction is important because uniform convergence preserves continuity, meaning the limit function will also be continuous if the original sequence of functions are continuous. This allows the interchange of limits and integrals or derivatives, which is often not valid for pointwise convergence. In the context of the exercise, uniform convergence of \( \sum (-1)^n \sin(a_n x) \) indicates that this series behaves consistently across any finite interval regardless of changes in \(x\). For uniform convergence, tools like the Weierstrass M-test become indispensable to assess whether a given function series meets this criterion.

• Uniform convergence results in a continuous limit if the original functions are continuous.
• It allows for the interchange of limit operations with integration and differentiation.
• It provides a stronger form of convergence than pointwise convergence.

Weierstrass M-test

The Weierstrass M-test is a practical tool used to determine the uniform convergence of a series of functions. This test provides a condition under which we can say a series converges uniformly by comparing it to a series of constant terms, called the comparison series.

To use the Weierstrass M-test, first identify a sequence \(M_n\) such that the absolute value of each function \(f_n(x)\) in the series is less than or equal to \(M_n\) for all \(x\) in the domain. That is, \(|f_n(x)| \leq M_n\). If the series \(\sum M_n\) converges, then the function series \(\sum f_n(x)\) also converges and does so uniformly over its domain.

In the exercise, we found \(M_n = a_n^{k+1}\) with the property that \(|(-1)^n \sin(a_n x)|\) is bounded by \(a_n^{k+1}\). Since the series \(\sum a_n^{k+1}\) converges, according to the Weierstrass M-test, the original series \(\sum (-1)^n \sin(a_n x)\) converges uniformly.

• The Weierstrass M-test is straightforward—it requires identifying a majorant series \(\sum M_n\) that controls the function series.
• If the comparison series \(\sum M_n\) converges, the function series converges uniformly.
• It effectively simplifies the analysis by focusing on a simpler series that dominates the terms of the function series.

Comparison Test

The comparison test is a fundamental technique for determining the convergence of series. It works by comparing the series of interest to another series that is already known to be convergent or divergent, depending on the inequality.

In the exercise, we use the comparison test to prove that \(\sum a_n^{k+1}\) converges. Given that \(a_{n+1} < a_n\) and the convergence of \(\sum a_n^k\), we found that \(0 \leq a_n^{k+1} \leq a_n^k\). Since \(\sum a_n^k < \infty\), the downward behavior and positive nature of \(a_n^{k+1}\) imply that this new series also converges due to the comparison test.

• Use the comparison test to establish convergence by relating the series to one you already know about.
• It requires finding a bounding sequence which is known to converge (or diverge).
• This method simplifies solving convergence problems by using inequalities and known series behavior.