Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The series diverges.

Step 1: Identify the General Term

First, let's identify what each term in the series looks like. The series is given by: \[\sum_{n=2}^{\infty} \left( \frac{1}{\sqrt{n}-1} - \frac{1}{\sqrt{n}+1} \right)\]This is the series we must analyze for convergence or divergence.

Step 2: Simplify the General Term

Simplify the expression for the general term:\[\frac{1}{\sqrt{n}-1} - \frac{1}{\sqrt{n}+1} = \frac{(\sqrt{n}+1) - (\sqrt{n}-1)}{(\sqrt{n}-1)(\sqrt{n}+1)}\]\[= \frac{2}{n-1}\]So, the general term simplifies to \( \frac{2}{n-1} \).

Step 3: Examine the Simplified Series

The series then becomes:\[2 \sum_{n=2}^{\infty} \frac{1}{n-1} = 2 \sum_{m=1}^{\infty} \frac{1}{m}\]Where we made the substitution \( m = n-1 \). This is a version of the harmonic series.

Step 4: Determine Convergence or Divergence

We know that the harmonic series, \( \sum_{m=1}^{\infty} \frac{1}{m} \), diverges. Since our series is essentially twice the harmonic series:\[2 \sum_{m=1}^{\infty} \frac{1}{m},\] it also diverges.

Key Concepts

Harmonic Series

The harmonic series is a simple, yet fundamental, series in mathematics. It has the form \(\sum_{n=1}^{\infty} \frac{1}{n} .\) Each term in the harmonic series is the reciprocal of a positive integer. Although these terms get smaller as \( n \) increases, they do not shrink fast enough to make the series converge.
In terms of behavior, the harmonic series is famous for its divergence. Unlike a convergent series, where the sum approaches a specific number, the harmonic series grows without bound. As you include more terms, the sum of the harmonic series increases indefinitely. Understanding the harmonic series is crucial when analyzing other series that resemble it, such as the altered series in this exercise.

General Term Simplification

General term simplification is a technique used to make complex series more manageable and easier to analyze. In this problem, each term in the series is expressed as:\(\frac{1}{\sqrt{n}-1} - \frac{1}{\sqrt{n}+1}.\) To simplify, we combine these fractions over a common denominator:

• Recognize that both terms share a similar structure but opposite denominators.
• This structure allows us to simplify using a difference of squares identity: \( (a-b)(a+b) = a^2 - b^2 \).

Applying this formula, the general term simplifies to \( \frac{2}{n-1} \). This simplification reveals that our unfamiliar series can be transformed into a format resembling the harmonic series, making further analysis more straightforward.

Divergence Analysis

Divergence analysis is an essential step in understanding whether a series sums to a finite value or grows indefinitely. In this exercise, once the general term is simplified to \( \frac{2}{n-1} \), we reframe the series as:\(2 \sum_{m=1}^{\infty} \frac{1}{m}\)where we made a substitution to align with the common harmonic series structure.
The task now is to determine if this series converges. Given the properties of the harmonic series, we know that it diverges, as its partial sums increase without limit. Therefore, multiplying by a constant, such as 2 in this case, does not alter the fundamental characteristic of infinite growth.

• If the base harmonic series \( \sum_{m=1}^{\infty} \frac{1}{m} \) diverges, the scaled version \( 2 \sum_{m=1}^{\infty} \frac{1}{m} \) also diverges.
• Divergence signifies that no finite sum can describe this series as it continues towards infinity.

Remember, recognizing divergence early can save time and effort in analyzing series behaviors.