Problem 18 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The series converges conditionally, but not absolutely.

Step 1: Understand the Series

We have the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n}}{1+\sqrt{n}} \). This is an alternating series because of the \((-1)^{n}\) factor, which causes the terms to alternate in sign.

Step 2: Apply the Alternating Series Test

The Alternating Series Test states that an alternating series \( \sum (-1)^{n} a_n \) converges if the sequence \( a_n \) is positive, decreasing, and \( \lim_{n \to \infty} a_n = 0 \).Here, \( a_n = \frac{1}{1+\sqrt{n}} \). We check:1. \( a_n > 0 \) for all \( n \), since 1 and \( \sqrt{n} \) are always positive.2. \( a_n \to 0 \) as \( n \to \infty \) because \( \sqrt{n} \) dominates as \( n \) grows.3. \( a_n \) is decreasing, since both the numerator is constant and the denominator \( 1+\sqrt{n} \) increases.Thus, the series converges by the Alternating Series Test.

Step 3: Test for Absolute Convergence

A series \( \sum a_n \) converges absolutely if \( \sum |a_n| \) converges. Here, \( |a_n| = \frac{1}{1+\sqrt{n}} \) for all \( n \).To test for absolute convergence, compare the series \( \sum \frac{1}{1+\sqrt{n}} \) to \( \sum \frac{1}{\sqrt{n}} \), which is a p-series with \( p = \frac{1}{2} \), known to diverge.Since \( \frac{1}{1+\sqrt{n}} \approx \frac{1}{\sqrt{n}} \) as \( n \to \infty \), and \( \frac{1}{\sqrt{n}} \) diverges, \( \sum \frac{1}{1+\sqrt{n}} \) also diverges by the Limit Comparison Test.

Key Concepts

Absolute Convergence

When we talk about absolute convergence, we're considering whether a series converges when we take the absolute value of every term. For a series \( \sum a_n \), absolute convergence occurs if the series \( \sum |a_n| \) converges. In simpler terms, regardless of the sign or direction of the terms, if their absolute magnitudes add up to a finite number, the series is absolutely convergent.

In our example, the series is \( \sum \frac{(-1)^{n}}{1+\sqrt{n}} \). To check for absolute convergence, we consider \( \sum \frac{1}{1+\sqrt{n}} \). If this series could be shown to converge, then the original series would converge absolutely. However, comparing \( \frac{1}{1+\sqrt{n}} \) to the classic divergent harmonic series \( \sum \frac{1}{\sqrt{n}} \) (a p-series with \( p=1/2 \), which is known to diverge), we find that \( \sum \frac{1}{1+\sqrt{n}} \) also diverges.

Therefore, although the original series converges, it does not converge absolutely, since its absolute version does not converge.

Convergence and Divergence

Convergence refers to a series approaching a specific finite value as we add more and more terms. On the other hand, divergence is when a series doesn't settle on any finite number; it might go to infinity or oscillate wildly.

For the series \( \sum \frac{(-1)^{n}}{1+\sqrt{n}} \), we assessed convergence using the Alternating Series Test. Alternating series are defined by terms that switch signs, often indicated by factors like \((-1)^{n}\). The series converges if the absolute terms of the series \( a_n = \frac{1}{1+\sqrt{n}} \) meet three criteria:

• The terms are positive (which they are since both 1 and \( \sqrt{n} \) are positive),
• The terms decrease (as \( n \to \infty \), \( \sqrt{n} \) grows, making the denominator larger, hence \( a_n \) is smaller),
• The terms converge to 0 as \( n \to \infty \).

These conditions being satisfied means the series converges.

However, as we've seen, convergence does not imply absolute convergence. While \( \sum \frac{(-1)^{n}}{1+\sqrt{n}} \) converges by alternating series test, it fails the absolute test.

Limit Comparison Test

The Limit Comparison Test is a handy tool for determining convergence or divergence of series when direct tests are challenging. By comparing a series to another, typically simpler series, we draw conclusions about the behavior of the original series.

To use the Limit Comparison Test, we identify two series, usually \( \sum b_n \) (the series under consideration) and \( \sum c_n \) (a known series). If \( \lim_{n \to \infty} \frac{b_n}{c_n} = L \), where \( L \) is a positive, finite number, both series either converge or diverge.

For the sequence \( \sum \frac{1}{1+\sqrt{n}} \), it's insightful to compare it with the harmonic-like series \( \sum \frac{1}{\sqrt{n}} \), known to diverge (since \( p=1/2 \)). As \( n \to \infty \), the term \( \frac{1}{1+\sqrt{n}} \) closely resembles \( \frac{1}{\sqrt{n}} \), suggesting that from the similarity, the series \( \sum \frac{1}{1+\sqrt{n}} \) diverges. The limit comparison result which matches the behavior shows both the compared series and the original series diverge.

This strategy simplifies analysis since it leverages familiar series with known convergence properties, making the evaluation process more intuitive.