Problem 35 - Thomas Calculus

Question

Which of the series Converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{\cos n \pi}{n \sqrt{n}} $$

Step-by-Step Solution

Answer

The series converges absolutely.

Step 1: Understanding Absolute Convergence

To determine absolute convergence, we first examine the series \( \sum_{n=1}^{\infty} \left| \frac{\cos n \pi}{n \sqrt{n}} \right| = \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \). A series converges absolutely if its corresponding series of absolute values converges.

Step 2: Test for Absolute Convergence Using P-Series

We recognize the series \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \) as a p-series where \( p = 3/2 > 1 \). A p-series converges if \( p > 1 \). Therefore, this series converges.

Step 3: Analyzing the Original Series for Convergence

The original series is \( \sum_{n=1}^{\infty} \frac{\cos n \pi}{n \sqrt{n}} \). We note that \( \cos n \pi = (-1)^n \), so the series can be rewritten as \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n^{3/2}} \), which is an alternating series.

Step 4: Apply the Alternating Series Test

To determine if the alternating series converges, use the Alternating Series Test. For this test, the terms \( b_n = \frac{1}{n^{3/2}} \) must be decreasing and \( \lim_{n \to \infty} b_n = 0 \). Since both conditions are met, the series converges.

Step 5: Conclusion on Type of Convergence

Since the series of absolute values converges, the original series converges absolutely. Therefore, the series \( \sum_{n=1}^{\infty} \frac{\cos n \pi}{n \sqrt{n}} \) converges absolutely and thus also converges.

Key Concepts

Absolute Convergence

To explore series convergence, we begin with the concept of absolute convergence. It plays a fundamental role in understanding series behavior. Imagine a series where you take each term and substitute its value with the absolute version—removing any negative signs. If this new series converges, we say that the original series converges absolutely. For example, the series \(\sum_{n=1}^{\infty} \left| \frac{\cos n \pi}{n \sqrt{n}} \right|\) transforms into \(\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}\). When the sum of the absolute values converges, the original series does too, absolutely. Absolute convergence suggests safety; in other words, regardless of the order of terms, the sum of the series remains the same. It's a stronger form of convergence, implying that the series converges without worrying about term signs.

P-Series Test

The p-series test is a valuable tool for assessing the convergence of series, particularly those resembling sums of the form \(\sum_{n=1}^{\infty} \frac{1}{n^p}\), where \(p\) is a positive number.If \(p > 1\), the series will converge. This happens because larger powers of \(n\) in the denominator makes the terms smaller as \(n\) increases, leading to a finite sum.In the example series \(\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}\), we see that \(p = 3/2\) which is greater than 1. Hence, the series converges. The beauty of the p-series test lies in its simplicity. It gives a clear and quick way to determine whether certain types of series will add up to a finite number, based merely on the exponent \(p\). This was incredibly useful in the example, providing proof of absolute convergence.

Alternating Series Test

The alternating series test is another convergence tool, primarily for series that alternate in sign. Such series look like \( \sum_{n=1}^{\infty} (-1)^n b_n \), where \(b_n\) are positive terms.Two things must happen for convergence here:

• The terms \(b_n\) must be decreasing—that means each term is smaller than the one before it as \(n\) increases.
• The limit of \(b_n\) as \(n\) approaches infinity must be zero—\( \lim_{n \to \infty} b_n = 0 \).

For the series \(\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{3/2}}\), these conditions are satisfied:1. The terms \(\frac{1}{n^{3/2}}\) clearly decrease as \(n\) increases.2. The limit here is zero, because dividing 1 by a large number results in a vanishingly small quantity.Therefore, the series converges. The alternating series test provided a second confirmation of convergence for the original series, ensuring the end result is valid.