Problem 44 - Thomas Calculus

Question

Which of the series in Exercises \(15 - 48\) converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { \sqrt { n } + \sqrt { n + 1 } } $$

Step-by-Step Solution

Answer

The series converges conditionally but not absolutely.

Step 1: Identify the Series Type

The given series is \(\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { \sqrt { n } + \sqrt { n + 1 } }\). It is an alternating series because the terms include \((-1)^n\), causing signs to alternate between positive and negative.

Step 2: Check for Absolute Convergence

For absolute convergence, evaluate the series \(\sum _ { n = 1 } ^ { \infty } \left| \frac { ( - 1 ) ^ { n } } { \sqrt { n } + \sqrt { n + 1 } } \right| = \sum _ { n = 1 } ^ { \infty } \frac { 1 } { \sqrt { n } + \sqrt { n + 1 } }\). Since \(\sqrt{n} + \sqrt{n+1} \approx 2\sqrt{n}\) for large \(n\), the terms behave like \(\frac{1}{2\sqrt{n}}\), which is a p-series with \(p = \frac{1}{2} < 1\). Therefore, this series diverges.

Step 3: Apply the Alternating Series Test

The alternating series test says that an alternating series \(\sum (-1)^n b_n\) converges if \(b_n\) is positive, decreasing, and \(\lim_{n \to \infty} b_n = 0\). Here, \(b_n = \frac{1}{\sqrt{n} + \sqrt{n+1}}\) is positive and \(\frac{1}{\sqrt{n} + \sqrt{n+1}}\) decreases as \(n\) increases. Also, \(\lim_{n \to \infty} \frac{1}{\sqrt{n} + \sqrt{n+1}} = 0\). Therefore, the series converges by the alternating series test.

Key Concepts

Alternating Series Test

The alternating series test is a widely used tool for determining the convergence of a series that alternates in sign. An alternating series takes the form \(\sum (-1)^n b_n\) where the terms \(b_n\) alternate between positive and negative as indicated by the factor \((-1)^n\). To decide if such a series converges, the following conditions must be met:

• \(b_n\) is positive for all \(n\).
• \(b_n\) is a decreasing sequence, which means each term is smaller than the previous term.
• The limit as \(n\) approaches infinity of \(b_n\) is zero, i.e., \(\lim_{n \to \infty} b_n = 0\).

If all these conditions are satisfied, the series converges. This test provides a simple method to verify convergence without necessarily finding the sum of the series, which is particularly useful in practical scenarios. In the given problem, because the sequence \(b_n = \frac{1}{\sqrt{n} + \sqrt{n+1}}\) meets these criteria, the series converges.

Absolute Convergence

Absolute convergence is a stronger condition than simple convergence. A series \(\sum a_n\) is said to converge absolutely if the series of its absolute values \(\sum |a_n|\) also converges. Mathematically, this means evaluating the series \(\sum |a_n|\) and checking through standard convergence tests if it converges.
To check for absolute convergence, consider the example given in the problem. The absolute value of the alternating series is \(\sum \left| \frac{(-1)^n}{\sqrt{n} + \sqrt{n+1}} \right| = \sum \frac{1}{\sqrt{n} + \sqrt{n+1}}\). If this series converges, the original series converges absolutely. However, in this example, it was found that the series behaves like a divergent \(p\)-series with \(p=1/2\). Hence, the series does not converge absolutely.

P-series

A \(p\)-series is a specific type of infinite series of the form \(\sum \frac{1}{n^p}\), where \(p\) is a constant. The convergence of a \(p\)-series depends entirely on the value of \(p\):

• If \(p > 1\), the series converges.
• If \(p \leq 1\), the series diverges.

This is based on the integral test for convergence, where the integral \(\int_1^\infty \frac{1}{x^p} \, dx\) converges for \(p > 1\). In the exercise case, the series is modified to \(\sum \frac{1}{2\sqrt{n}}\), effectively a \(p\)-series with \(p=1/2\). Since \(p=1/2 < 1\), the series is divergent according to the \(p\)-series test. This shows the importance of identifying and testing \(p\)-series for insights into the convergence behavior of certain kinds of series.

Divergent Series

A series is termed divergent if it does not converge. Divergence means that as more terms are added, the series does not approach a finite value; instead, it either grows indefinitely or oscillates. Divergent series may not satisfy certain convergence tests or criteria.
Regarding the problem series \(\sum \frac{1}{\sqrt{n} + \sqrt{n+1}}\), note that when considered without alternating signs (for absolute convergence), it behaves like the divergent \(p\)-series with \(p=1/2\). Thus, it diverges.
Understanding what makes a series diverge helps in identifying it and helps in the assessment of series convergence more broadly. Divergent series can sometime still be analyzed for other properties, but they won't settle onto a fixed sum like a convergent series would.