Problem 40 - Single Variable Calculus

Question

Show that the series \(\sum_{n=1}^{\infty} \frac{\sin \left(n^{3} x\right)}{n^{2}}\) converges for all values of \(x\), but \(\sum_{n=1}^{\infty} \frac{d}{d x}\left[\frac{\sin \left(n^{3} x\right)}{n^{2}}\right]\) diverges for all values of \(x\). Does this contradict Theorem 2? Explain your answer.

Step-by-Step Solution

Answer

The given series \(\sum_{n=1}^{\infty} \frac{\sin \left(n^{3} x\right)}{n^{2}}\) converges for all values of \(x\) by Dirichlet's test. However, its term-by-term derivative \(\sum_{n=1}^{\infty} n \cos(n^3 x)\) diverges for all values of \(x\) according to the Weierstrass M-test. This does not contradict Theorem 2 because we only showed the original series converges, not converges absolutely, and Theorem 2 applies to absolute convergence.

Step 1: Analyze the given series

We are given the series \(\sum_{n=1}^{\infty} \frac{\sin \left(n^{3} x\right)}{n^{2}}\). This series can be analyzed using the Weierstrass M-test, which states that if for each term in the series \(|a_n(x)| \leq M_n\) and the series \(\sum_{n=1}^{\infty} M_n\) converges, then the given series converges uniformly. We can use the inequality \(|\sin(\alpha)| \leq |\alpha|\) for any \(\alpha\), so we have: \[|\sin(n^3 x)| \leq |n^3 x|\] Thus \(|a_n(x)| \leq \frac{n^3 |x|}{n^2} = n|x|\). The series \(\sum_{n=1}^{\infty} n |x|\) diverges for all values of \(x\), so the Weierstrass M-test is inconclusive in this case.

Step 2: Prove the given series converges

From the previous analysis, we couldn't show the convergence of the given series using the Weierstrass M-test. However, we can use the Dirichlet's test, which states that a series \(\sum_{n=1}^{\infty} a_n b_n\) converges if the following conditions are met: 1. \(\{a_n\}\) is a sequence of real numbers which decreases monotonically to zero. 2. The sequence of partial sums of \(\{b_n\}\) is bounded. So we can rewrite our series as \(\sum_{n=1}^{\infty} \frac{1}{n^2} \sin(n^3 x)\). Here, \(a_n = \frac{1}{n^2}\) and \(b_n = \sin(n^3 x)\). It can be seen that the sequence \(a_n = \frac{1}{n^2}\) decreases monotonically to zero, satisfying the first condition of Dirichlet's test. The second condition requires the partial sums of \(\{b_n\}\) to be bounded. We can rewrite \(b_n = \sin(n^3 x)\) as a sum of two sinusoidal functions, using the trigonometric identity \(\sin(a+b) = \sin(a) \cos(b) + \cos(a)\sin(b)\): \[\sin(n^3 x) = \sin(n^3 x) - \sin((n-1)^3 x) + \sin((n-1)^3 x)\] As the sum of two sinusoidal functions with different frequencies is also a sinusoidal function with a bounded amplitude, it follows that the sequence of partial sums of \(b_n\) is bounded. Therefore, the series converges for all values of \(x\) by Dirichlet's test.

Step 3: Find the term-by-term derivative of the series

Now we find the term-by-term derivative of the series: \[\frac{d}{d x}\left[\frac{\sin \left(n^{3} x\right)}{n^{2}}\right] = \frac{n^3 \cos(n^3 x)}{n^2} = n \cos(n^3 x)\]

Step 4: Show that the term-by-term derivative diverges

To show that the term-by-term derivative diverges, we can use the Weierstrass M-test. For all \(x\), we have: \[|n \cos(n^3 x)| \leq n\] The series \(\sum_{n=1}^{\infty} n\) diverges, so by the Weierstrass M-test, the term-by-term derivative of the series \(\sum_{n=1}^{\infty} n \cos(n^3 x)\) diverges for all values of \(x\).

Step 5: Discuss whether this contradicts Theorem 2

Theorem 2 states that if a series converges absolutely, then the series formed by term-by-term differentiation also converges absolutely. In this case, the given series does converge for all values of \(x\), but its term-by-term derivative does not. However, we did not show that the original series converges absolutely, only that it converges. Thus, there is no contradiction with Theorem 2.

Key Concepts

Weierstrass M-test

The Weierstrass M-test is a valuable tool for determining if a series of functions converges uniformly. Here's how it works: you establish an upper limit or bound, known as \(M_n\), for each term of your series \(|a_n(x)|\). If the series \(\sum_{n=1}^{\infty} M_n\) converges, then the initial series also converges uniformly.

• You need each term \(|a_n(x)|\) to be less than or equal to \(M_n\).
• Then determine if the sum of all \(M_n\) values converges.

In our given problem, the series \(\sum_{n=1}^{\infty} \frac{\sin(n^3 x)}{n^2}\) was analyzed using the Weierstrass M-test. By observing that \(|\sin(n^3 x)| \leq |n^3 x|\), we determined that \(|a_n(x)| \leq n|x|\). Thus, the series \(\sum_{n=1}^{\infty} n|x|\) does not converge, making the Weierstrass M-test inconclusive in this situation.

Dirichlet's Test

Dirichlet's test is another way to determine if a series converges. It is useful when the Weierstrass M-test doesn't provide an answer. Dirichlet's test checks two key conditions:

• The sequence \(\{a_n\}\) must decrease monotonically to zero.
• The partial sums of the sequence \(\{b_n\}\) must be bounded.

In our exercise, the series is expressed as \(\sum_{n=1}^{\infty} \frac{1}{n^2} \sin(n^3 x)\). Here, \(a_n = \frac{1}{n^2}\) decreases monotonically to zero, and the sequence of partial sums of \(b_n = \sin(n^3 x)\) is bounded, as summed periodic functions remain within bounds. As both conditions are met, the series converges for all values of \(x\). This demonstrates the power of Dirichlet's test in proving convergence where other methods might fail.

Term-By-Term Differentiation

Term-by-term differentiation involves differentiating each individual term in a series and forming a new series from these derivatives. Simply put, you differentiate each term separately and create a series from these differentiations.

• This process assumes certain conditions about convergence to remain valid.
• It helps analyze the convergence of transformed series or derivative forms of functions.

For our specific series \(\sum_{n=1}^{\infty} \frac{\sin (n^{3} x)}{n^{2}}\), differentiating term-by-term gives us \(\sum_{n=1}^{\infty} n \cos(n^3 x)\). However, this series diverges, showing that careful consideration is required when applying term-by-term differentiation to determine convergence.

Absolute Convergence

Absolute convergence of a series means that the series remains convergent when we take the absolute value of each term. Mathematically, a series \(\sum_{n=1}^{\infty} a_n\) converges absolutely if \(\sum_{n=1}^{\infty} |a_n|\) converges.

• Absolute convergence guarantees regular convergence, but not the other way around.
• It ensures the series behaves well under operations like rearrangement and term-by-term differentiation.

In our exercise, though the original series \(\sum_{n=1}^{\infty} \frac{\sin (n^{3} x)}{n^{2}}\) converges, it does not converge absolutely as \(\sum_{n=1}^{\infty} \frac{1}{n^2} \) without the \(\sin\) function may suggest absolute convergence, yet we must include the periodic nature to confirm this. Thus, since absolute convergence wasn't established, term-by-term differentiation could lead to divergence, explaining why the differentiated series \(\sum_{n=1}^{\infty} n \cos(n^3 x)\) does not converge.