Problem 18 - Thomas Calculus

Question

Which of the series in Exercises \(15 - 48\) 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 given series converges, but not absolutely.

Step 1: Identify the Test for Convergence

The given series is \( \sum_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}} \). It is an alternating series because it has the term \((-1)^n\), which indicates alternating signs.

Step 2: Apply the Alternating Series Test

The Alternating Series Test states that an alternating series \( \sum (-1)^n a_n \) converges if: 1) \( a_n > 0 \) for all \( n \), 2) the sequence \( a_n \) is decreasing, and 3) \( \lim_{n \to \infty} a_n = 0 \). Here, \( a_n = \frac{1}{1+\sqrt{n}} \).

Step 3: Verify \( a_n > 0 \)

Since \( a_n = \frac{1}{1+\sqrt{n}} \) where both the numerator is positive and the denominator is positive for all \( n \geq 1 \), we conclude that \( a_n > 0 \) for all \( n \).

Step 4: Check if \( a_n \) is Decreasing

We need to show that \( a_n > a_{n+1} \). We simplify \( a_n = \frac{1}{1+\sqrt{n}} \) and \( a_{n+1} = \frac{1}{1+\sqrt{n+1}} \). Since \( \sqrt{n} < \sqrt{n+1} \), it follows that \( a_n > a_{n+1} \), making it a decreasing sequence.

Step 5: Check the Limit Condition

Calculate \( \lim_{n \to \infty} \frac{1}{1+\sqrt{n}} \). As \( n \rightarrow \infty \), \( \sqrt{n} \rightarrow \infty \) which implies \( 1 + \sqrt{n} \rightarrow \infty \). Therefore, \( \lim_{n \to \infty} \frac{1}{1+\sqrt{n}} = 0 \).

Step 6: Conclude with the Alternating Series Test

Since all conditions of the Alternating Series Test are satisfied – \( a_n > 0 \), \( a_n \) is decreasing, and \( \lim_{n \to \infty} a_n = 0 \) – the series converges by the Alternating Series Test.

Step 7: Absolute Convergence Check

To check for absolute convergence, consider \( \sum_{n=1}^{\infty} \left| \frac{(-1)^n}{1+\sqrt{n}} \right| = \sum_{n=1}^{\infty} \frac{1}{1+\sqrt{n}} \). Since this series resembles the \( p \)-series \( \sum_{n=1}^{\infty} \frac{1}{n^{1/2}} \), which diverges, the series does not converge absolutely.

Key Concepts

Absolute Convergence

Absolute convergence is a stronger form of convergence for an infinite series. If a series is absolutely convergent, then not only does the series converge, but the series formed by taking the absolute values of its terms also converges.
To check for absolute convergence in the given series, we remove the alternating factor by considering the series \( \sum_{n=1}^{\infty} \left| \frac{(-1)^n}{1+\sqrt{n}} \right| = \sum_{n=1}^{\infty} \frac{1}{1+\sqrt{n}} \).
This series resembles the \( p \)-series \( \sum_{n=1}^{\infty} \frac{1}{n^{1/2}} \), which is known to diverge when \( p \leq 1 \). Therefore, our series does not converge absolutely.
This means, despite the series being convergent, its absolute counterpart fails to meet the convergence criteria.

Series Convergence

Convergence of a series means that the series approaches a finite value as the number of its terms increases indefinitely. For our series \( \sum_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}} \), we applied the Alternating Series Test to determine convergence.
Key points:

• The terms \( a_n = \frac{1}{1+\sqrt{n}} \) must be positive, which they are for all \( n \).
• The terms must form a decreasing sequence. Given that as \( n \) increases, \( \sqrt{n} \) grows, making \( a_n > a_{n+1} \), they satisfy this condition.
• The limit of the terms should approach zero as \( n \) approaches infinity. Here, \( \lim_{n \to \infty} \frac{1}{1+\sqrt{n}} = 0 \).

Since all these criteria are met, the series converges according to the Alternating Series Test.

Alternating Series

An alternating series is a series where the signs of the terms alternate between positive and negative. This occurs often with a factor of \((-1)^n\) or \((-1)^{n+1}\).
The series \( \sum_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}} \) is an example of an alternating series. The presence of \((-1)^n\) ensures the terms alternate signs as \( n \) progresses, creating a pattern of positive and negative terms.
For such series, we utilize the Alternating Series Test which checks that, with decreasing terms and limit approaching zero, the overall series converges.
This test plays a crucial role in simplifying series analysis when absolute convergence is not present.

Limit of a Sequence

Understanding the limit of a sequence is fundamental for determining the behavior of series. In the context of this problem, the sequence in question is \( a_n = \frac{1}{1+\sqrt{n}} \).
As \( n \rightarrow \infty \), it is important to examine \( \lim_{n \to \infty} a_n \). Here, as \( \sqrt{n} \rightarrow \infty \), \(1 + \sqrt{n} \) becomes very large. Consequently, \( a_n \) approaches 0.
Ensuring that the sequence approaches zero is vital for convergence in alternating series as stated by the Alternating Series Test.
Thus, determining the limit helps strengthen the argument that our series converges subject to the test’s requirements.