Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

Does \(\sum_{n=3}^{\infty} 1 /[n \cdot \ln n \cdot \ln (\ln n)]\) converge or diverge? Explain.

Step-by-Step Solution

Answer

The series diverges.

Step 1: Recognize the Type of Series

The given series is \( \sum_{n=3}^{\infty} \frac{1}{n \cdot \ln n \cdot \ln (\ln n)} \). This series is a type of generalized harmonic series.

Step 2: Apply the Cauchy Condensation Test

The Cauchy Condensation Test is a useful tool for series where terms are decreasing and positive. For a series \( \sum a_n \) with \( a_n = \frac{1}{n \cdot \ln n \cdot \ln (\ln n)} \), the Cauchy Condensation Test tells us to consider the series \( \sum 2^k a_{2^k} \).

Step 3: Simplify the Condensed Series

Calculate \( a_{2^k} = \frac{1}{2^k \cdot \ln (2^k) \cdot \ln (\ln (2^k))} \). We need to estimate \( \ln (2^k) = k \ln 2 \) and \( \ln (\ln (2^k)) = \ln (k \ln 2) \).

Step 4: Analyze the Behavior of Series

The condensed series becomes \( \sum \frac{2^k}{2^k k \ln 2 \cdot \ln (k \ln 2)} = \sum \frac{1}{k \ln 2 \cdot \ln (k \ln 2)} \), which resembles the Harmonic series. Compare it with \( \sum \frac{1}{k \cdot \ln k} \), known to diverge.

Step 5: Conclusion on Convergence or Divergence

Since \( \sum \frac{1}{k \ln 2 \cdot \ln (k \ln 2)} \) is similar in form to a diverging series, it also diverges. Consequently, the original series \( \sum_{n=3}^{\infty} \frac{1}{n \cdot \ln n \cdot \ln (\ln n)} \) diverges as well.

Key Concepts

Cauchy Condensation Test

The Cauchy Condensation Test is a powerful tool when it comes to understanding the behavior of infinite series, especially those consisting of positive and decreasing terms. This test helps us determine whether such a series converges or diverges.

Here's how it works: if you have a series \( \sum a_n \) where each \( a_n \) is positive and decreasing, the Cauchy Condensation Test suggests you consider the series \( \sum 2^k a_{2^k} \). The gist of it is to replace each term in the series with a new term, where the index is doubled. The condensed series resembles a simpler format that can be easier to evaluate for convergence or divergence.

• If the condensed series converges, so does the original series.
• If the condensed series diverges, the original series also diverges.

By simplifying the problem, the Cauchy Condensation Test provides a new lens to view the convergence behavior. In this example, we applied it to certain terms involving natural logarithms to understand how they grow and behave. This simplification helped us reach our final conclusion about divergence much easier.

Harmonic Series

The Harmonic Series is one of the most famous examples of a divergent series. It is given by the sum \( \sum_{n=1}^{\infty} \frac{1}{n} \). Despite the terms getting exceedingly small, this series diverges. In other words, as you add an infinite number of these diminishing terms, the total sum grows without bound.

The behavior of the Harmonic Series is essential to understand other series that might resemble it somewhat, even if only in a generalized form. The Harmonic Series diverges, which means that if you find a series similar in form, be cautious – it might diverge too!

• It is an archetype of comparison when analyzing other series.
• Interactions with logarithmic factors often lead to intriguing results.

Understanding its nature allows us to predict the behavior of related series. It is a crucial concept when looking at series with factors like \( \ln n \) in their terms. These can form what is known as generalized harmonic series.

Generalized Harmonic Series

A Generalized Harmonic Series extends the concept of the classic harmonic series by including additional factors or changing its structure in certain ways. For instance, a common form is \( \sum \frac{1}{n^p} \) or others like \( \sum \frac{1}{n \ln n} \). These variations change the convergence properties of the series.

The important parameter here is \( p \). In the series \( \sum \frac{1}{n^p} \),

• If \( p > 1 \), the series converges.
• If \( p \leq 1 \), the series diverges.

Generalized harmonic series with logarithmic components, such as \( \sum \frac{1}{n \ln n} \), are known to diverge. Adding to this is the complexity of even more logarithmic factors, as in the original exercise, making them prime candidates for the Cauchy Condensation Test. By transforming the original series, we can more quickly understand its convergence behavior by relating it to known forms.

Divergence of Series

The divergence of a series implies that as you add up its terms indefinitely, the sum grows without limit. In the world of calculus and analysis, differentiating between converging and diverging series is vital since it tells us if a series has a finite sum.

In the provided exercise, the series diverged due to a resemblance to known divergent forms like \( \sum \frac{1}{k \ln k} \). When analyzing such series, look for familiar patterns or simplify via tests such as the Cauchy Condensation Test.

• Divergent series often seem to shrink but still diverge due to certain key behaviors.
• Logarithmic terms especially require careful attention as they can diminish slowly.

Ultimately, recognizing the divergence of a series is crucial since it tells us about the potential infinity of the sum or the nature of the sequence involved. This understanding is foundational in calculus and informs decisions in theoretical and practical applications.