Problem 31 - Thomas Calculus

Question

Which of the series in Exercises \(17-56\) converge, and which diverge? Use any method, and give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{1}{1+\ln n} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{1}{1 + \ln n} \) diverges.

Step 1: Identify the Series Type

The series given is \( \sum_{n=1}^{\infty} \frac{1}{1 + \ln n} \). It is neither a simple geometric nor an arithmetic series. The expression involves a logarithm, indicating the use of a comparison test might be suitable.

Step 2: Consider the Comparison Test

To apply the comparison test, compare the given series to a known series. Notice that \( \ln n \to \infty \) as \( n \to \infty \), hence \( \frac{1}{1 + \ln n} \) behaves like \( \frac{1}{\ln n} \) for large \( n \).

Step 3: Compare with a Known Divergent Series

The harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \) is a known divergent series. Since for large \( n \), \( \ln n < n \), it follows that \( \frac{1}{\ln n} > \frac{1}{n} \). Therefore, \( \frac{1}{1+ \ln n} > \frac{1}{n+ \ln n} \) for large \( n \).

Step 4: Use Limit Comparison Test

For the limit comparison test between \( \frac{1}{1+ \ln n} \) and \( \frac{1}{n} \), calculate \( \lim_{n \to \infty} \frac{\frac{1}{1+ \ln n}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n}{1+ \ln n} \).

Step 5: Evaluate the Limit

Evaluate \( \lim_{n \to \infty} \frac{n}{1+ \ln n} \). As \( n \to \infty \), \( \ln n \to \infty \) more slowly than \( n \), thus \( \frac{n}{1+ \ln n} \to \infty \). The limit is infinite, indicating \( \sum_{n=1}^{\infty} \frac{1}{1+ \ln n} \) behaves like \( \sum_{n=1}^{\infty} n \), which diverges.

Step 6: Conclusion

Since \( \lim_{n \to \infty} \frac{n}{1+ \ln n} = \infty \), and given the divergence of the harmonic series, by the limit comparison test, \( \sum_{n=1}^{\infty} \frac{1}{1+ \ln n} \) diverges as well.

Key Concepts

Comparison Test

In the study of series, the comparison test is a valuable tool for determining the convergence or divergence of a series. The idea behind the comparison test is quite straightforward. We compare the series in question with another series whose behavior—either convergence or divergence—is already known. If the series being tested has smaller values compared to a known convergent series, then it also converges. Conversely, if it is larger than a known divergent series, it diverges.

To apply this test effectively, you should:

• Ensure that all terms in the series are positive.
• Find a series that closely resembles the one being tested and whose behavior (convergent/divergent) is known.
• Check the inequality relations between the terms of the series being compared.

Applying this test requires analytical skill to choose an appropriate comparison series and analyze the given limits.

Logarithmic Series

Logarithmic series involve terms containing logarithmic expressions, like the series we are considering: \( \sum \frac{1}{1+\ln n} \). These series can be challenging to categorize because their behavior falls between commonly understood patterns like geometric or arithmetic sequences. In this kind of series, we often turn to logarithms' inherent properties—such as slow growth at infinity—to gain insights.

When the terms of a series contain logarithms, they might resemble other series' behavior for large values of \( n \). An approach to solve these series often involves simplifying or approximating them using familiar series like the harmonic series. This simplification allows us to leverage known results and apply tests like the comparison test to derive conclusions.

Harmonic Series

The harmonic series \( \sum_{n=1}^{finity} \frac{1}{n} \) is a classic example in the study of divergent series. Despite its simple appearance, this series is known to diverge, meaning the sum grows without bound as more terms are added. The divergence of the harmonic series is an essential characteristic to understand because it helps to form the basis for comparing other series.

In practice, the harmonic series serves as a benchmark in the limit comparison test or the standard comparison test. Any series that, term by term, exceeds those of the harmonic series is guaranteed to diverge. This property makes the harmonic series a useful reference point when working with other series exhibiting specific logarithmic or polynomial behaviors.

Limit Comparison Test

The limit comparison test is another powerful method for analyzing the convergence of a series. This test involves taking the limit of the ratio between the given series' terms and those of a known series. The success of this test hinges on determining whether this ratio approaches a finite non-zero value or infinity.

To apply the limit comparison test:

• Choose a known convergent or divergent series to compare with.
• Compute the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \), where \( a_n \) are terms from the series being tested and \( b_n \) are from the comparison series.
• If the limit exists and is finite and non-zero, both series either converge or diverge together.

This test proves useful when direct comparisons with inequalities are challenging, offering a reliable way to draw conclusions about a series by closely analyzing term ratios.