Problem 13 - Calculus: Early Transcendentals

Question

Determine whether the series is convergent or divergent. \( \frac {1}{3} + \frac {1}{7} + \frac {1}{11} + \frac {1}{15} + \frac {1}{19} + \cdot \cdot \cdot \)

Step-by-Step Solution

Answer

The series is divergent.

Step 1: Identify the Pattern

Observe the denominators of the series: 3, 7, 11, 15, 19, ... These form an arithmetic sequence where the first term is 3, and the common difference, d, is 4. Thus, the general term for the denominator in the nth term is given by \(a_n = 4n - 1\).

Step 2: Formulate the nth Term of the Series

Given the pattern in the denominators, the nth term of the series is \(\frac{1}{4n - 1}\). Thus, the series can be expressed as \( \sum_{n=1}^{\infty} \frac{1}{4n - 1} \).

Step 3: Apply the Divergence Test

The Divergence Test states that if \(\lim_{n \to \infty} a_n eq 0\), then the series diverges. Apply this to our series: \(\lim_{n \to \infty} \frac{1}{4n - 1} = 0\), so the divergence test is inconclusive as the limit equals zero.

Step 4: Compare to a Known Series

We compare the given series to the harmonic series \(\sum_{n=1}^{\infty} \frac{1}{n}\), which diverges. Observe that for all \(n\), \(\frac{1}{4n - 1} > \frac{1}{4n}\). By the Comparison Test, since \(\sum_{n=1}^{\infty} \frac{1}{4n}\) diverges, our series \(\sum_{n=1}^{\infty} \frac{1}{4n - 1}\) also diverges.

Key Concepts

Divergence Test

The Divergence Test is a straightforward and useful tool for determining whether a series could be divergent. This test is one of the first you might consider because of its simplicity. It states that if the limit of the nth term of a series as n approaches infinity does not equal zero, the series is divergent. In other words, if \( \lim_{n \to \infty} a_n eq 0\), then \( \sum_{n=1}^{\infty} a_n \) diverges.

However, it's important to note that this test is only able to confirm divergence and not convergence. For instance, if you apply the divergence test to a series and find that \( \lim_{n \to \infty} a_n = 0 \), the test is inconclusive. This means the series could either converge or diverge. More advanced tests, like the Comparison Test, may be required in such cases to conclude whether the series is convergent or divergent.

Comparison Test

The Comparison Test is a valuable method used to determine the convergence or divergence of an infinite series by comparing it to another series whose behavior is already known. The idea is simple: compare your series to a benchmark series.- If you're trying to prove a series converges, compare it to a series you know converges. If each term in your series is less than the corresponding term in the known series, and the known series converges, your series converges too.- Conversely, if you wish to prove divergence, compare your series to one that diverges. If each term in your series is bigger than the corresponding term in a known divergent series, your series diverges.In the exercise, we compared the series \( \sum_{n=1}^{\infty} \frac{1}{4n - 1} \) with the harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \), a known divergent series. We observed that for all \( n \), \( \frac{1}{4n - 1} > \frac{1}{4n} \). We know \( \sum_{n=1}^{\infty} \frac{1}{4n} \) diverges (since it is a constant multiple of the harmonic series), leading us to conclude using the Comparison Test that \( \sum_{n=1}^{\infty} \frac{1}{4n - 1} \) diverges as well.

Arithmetic Sequence

An arithmetic sequence is a sequence of numbers where each term after the first is generated by adding a constant, called the common difference \(d\), to the previous term. The pattern needs to be identified to analyze series behavior effectively and create a mathematical representation of the series terms.In the original problem's series, the denominators 3, 7, 11, 15, and 19 form an arithmetic sequence:- The first term \(a_1\) is 3.- The common difference \(d\) is 4, calculated by subtracting the first term from the second term (\(7 - 3 = 4\)).Using this information, the nth term of the arithmetic sequence can be expressed as \( a_n = 4n - 1 \). This formula is crucial as it informs us of how the denominators behave and allows us to express the general term for the series as \( \frac{1}{4n - 1} \). Understanding this sequence foundation underpins our comprehension of the series' behavior—in this case, leading to a divergence conclusion.