Problem 16 - Single Variable Calculus

Question

Determine whether the series converges or diverges. $$ \sum_{n=2}^{\infty}(\ln n) \sin \frac{(2 n-1) \pi}{2} $$

Step-by-Step Solution

Answer

The given series is \(\sum_{n=2}^{\infty}(\ln n) \sin \frac{(2 n-1) \pi}{2}\). It is an alternating series due to the sine function. However, the alternating series test cannot be used to determine convergence or divergence, since the magnitude of the terms, \(|\ln n|\), is increasing and does not approach zero. There's a high chance that the series diverges. To reach a more formal conclusion, another test such as the Direct Comparison Test or Limit Comparison Test could be applied.

Step 1: Identify the alternating series

The given series can be written as: $$ \sum_{n=2}^{\infty}(\ln n) \sin \frac{(2 n-1) \pi}{2} $$ Notice that the sine function causes the terms to be alternating, as the argument increases by \(\frac{\pi}{2}\) for each term, causing the sine function to oscillate between positive and negative values.

Step 2: Check if the magnitude of the terms is non-increasing

We need to check if \(|b_n|\) is non-increasing. Let's find the derivative: $$ |b_n| = |\ln n| $$ Now, let us find the derivative of \(|\ln n|\) with respect to n: $$ \frac{d}{dn} |\ln n| = \frac{1}{n} $$ Since \(\frac{1}{n} > 0\) for all \(n \geq 2\), the magnitude of the terms, i.e., \(|\ln n|\), is increasing. Hence, the first condition of the alternating series test is not satisfied.

Step 3: Conclusion

Since the first condition for the alternating series test is not satisfied, we cannot conclude about the convergence or divergence of the series using this test. However, since the magnitude of the terms does not decrease or approach zero, there's a high chance that the series diverges. If a more formal conclusion is needed, another test such as the Direct Comparison Test or the Limit Comparison Test could be applied.

Key Concepts

Convergence Tests

Understanding whether a series converges or diverges is crucial in calculus, specifically in the study of infinite series. Convergence tests are a collection of techniques used to evaluate if an infinite series adds up to a finite number. Different tests can apply based on the series' characteristics.

• Alternating Series Test: Useful for series whose terms alternate in sign.
• Direct Comparison Test: Compares terms of the series to a known convergent or divergent series.
• Limit Comparison Test: Uses the limit of the ratio of terms to compare series.

When tackling a series, selecting an appropriate convergence test can simplify finding out whether a series converges, helping avoid misleading conclusions.

Divergence

Divergence occurs when the series does not sum to a finite number. Instead, the terms grow or oscillate without settling into a sum. Recognizing divergence can prevent wasted effort on finding an elusive sum. For example, in the given series,\[\sum_{n=2}^{\infty}(\ln n) \sin \frac{(2 n-1) \pi}{2},\]we suspect divergence as the series terms do not settle down due to increasing \(\ln n\). Tests like the Alternating Series Test could indicate this divergence, highlighting when terms remain large or inappropriate for convergence.

Alternating Series Test

The Alternating Series Test is specifically designed for series with terms that switch signs. Two main conditions must be satisfied:

• The absolute value of the terms must be decreasing, meaning \( |b_{n+1}| \leq |b_n| \) for all \(n\) beyond some point.
• The limit of the absolute terms approaching zero, i.e., \( \lim_{n \to \infty} b_n = 0 \).

In the series \[\sum_{n=2}^{\infty}(\ln n) \sin \frac{(2 n-1) \pi}{2},\]the alternating nature is defined by the sine function. However, since \( |\ln n| \) is increasing and not decreasing, this test cannot confirm convergence, showing why the magnitude of terms is crucial.

Limit Comparison Test

The Limit Comparison Test serves as a tool to determine convergence by comparing a complex series with a simpler one. It involves:

• Taking the limit of the ratio of the series' terms to a comparison series.
• If this limit exists and is finite and positive, both series will converge or diverge together.

For example, when direct application of the Alternating Series Test fails, as in the given series, this test can often step in, comparing against benchmark series like \(\sum \frac{1}{n^p} \) to derive conclusions about convergence behavior.

Direct Comparison Test

The Direct Comparison Test checks if a series can be compared directly to another known convergent or divergent series. It works by:

• Ensuring that terms are positive and comparable.
• Using a series where \( b_n \leq a_n \) indicates convergence if \sum b_n\ converges, and \( b_n \geq a_n \) indicates divergence if \sum b_n\ diverges.

In our series, comparing to \(\ln n\) might allow comparison with harmonic or p-series. While not applied here, when appropriate, it provides a straightforward method to conclude series behavior.