Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

Discuss the convergence or divergence of $$ \begin{array}{r} \frac{1}{\sqrt{2}-1}-\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}+ \\\ \frac{1}{\sqrt{4}-1}-\frac{1}{\sqrt{4}+1}+\cdots \end{array} $$

Step-by-Step Solution

Answer

The series diverges.

Step 1: Analyze the General Term

Consider the general term of the sequence as \( a_n = \frac{1}{\sqrt{n}-1} - \frac{1}{\sqrt{n}+1} \). We need to analyze whether the terms simplify.

Step 2: Simplify the General Term

Simplify the expression \( a_n = \frac{1}{\sqrt{n}-1} - \frac{1}{\sqrt{n}+1} \) by finding a common denominator:\[a_n = \frac{(\sqrt{n}+1) - (\sqrt{n}-1)}{(\sqrt{n}-1)(\sqrt{n}+1)} = \frac{2}{n-1}.\]

Step 3: Examine the Series' Terms

The series becomes \( \sum_{n=2}^{\infty} \frac{2}{n-1} = 2 \sum_{n=2}^{\infty} \frac{1}{n-1} \), which can be rewritten for clarity as \( 2 \sum_{m=1}^{\infty} \frac{1}{m} \).

Step 4: Determine Convergence or Divergence

The series \( \sum_{m=1}^{\infty} \frac{1}{m} \) is a harmonic series, which is well-known to diverge. Therefore, our original series \( 2 \sum_{m=1}^{\infty} \frac{1}{m} \) also diverges.

Key Concepts

Harmonic Series

The harmonic series is a fundamental concept when exploring the convergence or divergence of mathematical series. It is represented as \[ \sum_{n=1}^{\infty} \frac{1}{n} \]This series starts with the term \( \frac{1}{1} \) + \( \frac{1}{2} \) + \( \frac{1}{3} \) + \( \frac{1}{4} \) + \ldots
To understand convergence, we evaluate whether the sum of all terms has a finite limit as \( n \to \infty \). However, the harmonic series diverges.
The partial sums \( S_n = 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} \)increase without limit.

• This means that as we add more terms, the sum keeps growing.
• Although terms go towards zero, they don't decrease quickly enough for the series to converge.

In essence, the harmonic series gets indefinitely larger, making it diverge.

General Term

The general term of a series gives insight into its form and behavior. For the exercise in question, the general term is expressed as: \[a_n = \frac{1}{\sqrt{n}-1} - \frac{1}{\sqrt{n}+1} \]This expression needs simplification to better understand its contribution to the whole series.
When simplifying:

• Find a common denominator for the two parts.
• Perform algebraic operations: \[a_n = \frac{(\sqrt{n}+1) - (\sqrt{n}-1)}{(\sqrt{n}-1)(\sqrt{n}+1)} = \frac{2}{n-1} \]

Such manipulation reveals how each term behaves and is pivotal in analyzing the series' divergence or convergence.
Understanding the general term is crucial as it provides a clearer evaluation of the whole series.

Mathematical Sequence

A mathematical sequence is comprised of ordered terms. Sequences are foundational when discussing series. Our series is derived from the general term \(a_n = \frac{2}{n-1} \).
Sequences have a specific form:

• They can be finite, with a clear start and end.
• Or infinite, which continuously goes on.

The sequence related to the exercise is infinite. This is crucial because infinite sequences require careful analysis to understand convergence and divergence behavior.
The sequence \( \left( \frac{2}{1}, \frac{2}{2}, \frac{2}{3}, \ldots \right) \)influence the sum's outcome, as in our given example, it contributes to identifying the overall divergent nature of the series when summed.

Infinite Series

An infinite series is the sum of the terms of an infinite sequence. It plays a significant role in determining finite or infinite sums from these sequences. Our given series is expressed as:\[ \sum_{n=2}^{\infty} \frac{2}{n-1} \]This indicates an infinitely long sum starting from a particular term. Understanding the characteristics of such infinite series includes:

• Identifying the series' formation and limits.
• Determining its convergence or divergence by observing the general term and summation behavior.

In our case, after simplifying, the series leads to the well-known harmonic series:\[ 2 \sum_{n=1}^{\infty} \frac{1}{n} \]which is established to diverge. This implies that infinite series with similar configurations also tend to diverge unless their terms diminish rapidly enough.