Problem 12 - Calculus: Early Transcendentals

Question

Determine whether the series is convergent or divergent. \( \frac {1}{5} + \frac {1}{7} + \frac {1}{9} + \frac {1}{11} + \frac {1}{13} + \cdot \cdot \cdot \)

Step-by-Step Solution

Answer

The series diverges by the Comparison Test.

Step 1: Identify the Series Form

The given series can be written in the form \( \sum_{n=1}^{\infty} \frac{1}{2n+3} \). This is known as a harmonic-like series, where the denominator increases with each term.

Step 2: Comparison Test with a Known Series

Compare the series \( \sum_{n=1}^{\infty} \frac{1}{2n+3} \) with the divergent harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \). Note that for all \( n \geq 1, \frac{1}{2n+3} \geq \frac{1}{3n} \), and the series \( \sum_{n=1}^{\infty} \frac{1}{3n} \) diverges like the harmonic series.

Step 3: Apply the Comparison Test

Since \( \sum \frac{1}{3n} \) diverges, and \( \frac{1}{2n+3} \geq \frac{1}{3n} \), by the Comparison Test, \( \sum \frac{1}{2n+3} \) also diverges. The Comparison Test states that if \( \sum a_n \geq \sum b_n \) and \( \sum b_n \) is divergent, then \( \sum a_n \) is also divergent.

Key Concepts

Divergent Series

A divergent series is a series that does not have a finite sum. This means that as you keep adding the terms of the series, the total sum grows indefinitely. Series can diverge by going to positive or negative infinity, or even by fluctuating wildly without settling towards any particular value.

Understanding divergence is crucial, especially when working with infinite series. If a series diverges, it means there's no single number it converges to, no matter how many terms you add together. In technical terms, we say a series \( \sum a_n \) diverges if the sequence of its partial sums \( S_n = a_1 + a_2 + ... + a_n \) does not approach a finite limit as \( n \) becomes very large.

Comparison Test

The comparison test is a handy tool to determine the convergence or divergence of a series, especially when dealing with series that seem similar to others we already understand. The basic idea is to compare a series with terms \( a_n \) to another series with known behavior (either convergent or divergent).

• When using the comparison test, start by identifying a series \( \sum b_n \) that is either clearly convergent or divergent.
• If for all sufficiently large terms, your series has terms \( a_n \geq b_n \) and the series \( \sum b_n \) diverges, then \( \sum a_n \) also diverges.
• Conversely, if \( a_n \leq b_n \) and \( \sum b_n \) converges, then \( \sum a_n \) converges too.

In the original exercise, we compared our given series to a transformed harmonic series \( \sum \frac{1}{3n} \). Because the harmonic series diverges, if our series has larger terms than \( \sum \frac{1}{3n} \), it must also diverge.

Harmonic Series

One of the most famous examples of a divergent series is the harmonic series. The harmonic series is given by:\[\sum_{n=1}^{\infty} \frac{1}{n} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\]At first glance, this might seem like it would converge because each term is getting smaller. However, it actually diverges. As you keep adding more terms, the partial sums keep growing without bound.

• The basic idea is that even though terms are decreasing, they decrease too slowly to ever sum up to a finite limit.
• This property makes the harmonic series an excellent benchmark for applying the comparison test.
• If a series can be shown to be greater than or comparable to a harmonic series, it will also tend to diverge.

The exercise used the harmonic series as a reference point. By comparing the series in the problem to a related form of the harmonic series, we established its divergence.