Problem 27 - Calculus: Early Transcendentals

Question

Determine whether the series is convergent or divergent. If it is convergent, find its sum. \( \frac {1}{3} + \frac {1}{6} + \frac {1}{9} + \frac {1}{12} + \frac {1}{15} + \cdot \cdot \cdot \)

Step-by-Step Solution

Answer

The series is divergent.

Step 1: Identify the Series

The given sequence follows the pattern: \( \frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{12}, \frac{1}{15}, \ldots \). This series can be expressed in the general form: \( a_n = \frac{1}{3n} \).

Step 2: Recognize the Type of Series

Notice that the series has the general term \( a_n = \frac{1}{3n} \). It resembles the harmonic series \( \sum \frac{1}{n} \), which is a well-known divergent series.

Step 3: Divergence Test for the Series

The harmonic series \( \sum \frac{1}{n} \) is divergent, and since \( \sum \frac{1}{3n} \) is just \( \frac{1}{3} \) times the harmonic series, it is also divergent. Therefore, \( \sum \frac{1}{3n} = \frac{1}{3} \sum \frac{1}{n} \) is divergent.

Step 4: Conclusion

Since the series \( \sum \frac{1}{3n} \) has terms that form a harmonic-like series with a constant multiplier, and it is divergent, there is no finite sum. Thus, the series \( \frac{1}{3} + \frac{1}{6} + \cdots \) is divergent.

Key Concepts

Harmonic Series

Imagine standing in line, where each person stands a little bit farther away from the next. That's similar to how a harmonic series is structured. The harmonic series is expressed as \( \sum \frac{1}{n} \), where \( n \) begins at 1 and goes to infinity. Each term in this series is the reciprocal of an integer. There's a fascinating aspect to this series: many people might expect it to converge to a certain value, but surprisingly, it doesn't. The harmonic series grows indefinitely as you add more terms, which is why it is classified as divergent. This is a key concept to grasp, especially when determining the nature of other similar series.

Divergent Series

A divergent series is like a car accelerating infinitely without ever stopping; it never settles at a destination. Divergent series are those that do not approach a finite value, no matter how many terms you add. In mathematical terms, the sequence of partial sums does not converge to a fixed number. A classic example is the harmonic series, \( \sum \frac{1}{n} \), which we've established as divergent. If you multiply all its terms by a constant, such as \( \frac{1}{3} \), the new series \( \sum \frac{1}{3n} \) remains divergent. Hence, once a series is proven divergent, multiplying by a non-zero constant does not change its divergence.

General Term of a Series

The general term of a series, denoted as \( a_n \), is a blueprint for all the terms in that series. It is a formula that you can use to find any term's value in the sequence just by plugging in the position number \( n \). In the exercise you examined, the general term is \( a_n = \frac{1}{3n} \). This term helps quickly recognize patterns and identify the series type. For instance, when you saw \( \frac{1}{3}, \frac{1}{6}, \ldots \), realizing its general form \( \frac{1}{3n} \) allowed for easy identification with the harmonic series with a constant factor. Knowing \( a_n \) helps mathematicians and students alike figure out whether a series might be potentially convergent or divergent, and simplifies the computation of each term in the sequence.