Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that \(\sum_{k=1}^{\infty} \ln \frac{k}{k+1}\) diverges. Hint: Obtain a formula for \(S_{n}\).

Step-by-Step Solution

Answer

The series diverges because the partial sum \( S_n \to -\infty \) as \( n \to \infty \).

Step 1: Understand the Series

The series given is \( \sum_{k=1}^{\infty} \ln \frac{k}{k+1} \). We need to determine if this series converges or diverges.

Step 2: Express the Series

Notice that \( \ln \frac{k}{k+1} = \ln k - \ln(k+1) \). This form is beneficial for applying telescoping, as it consists of consecutive logarithmic differences.

Step 3: Recognize the Telescoping Nature

The terms in the series are \( \ln k - \ln(k+1) \). When expanded, this sequence allows cancellation of intermediate terms as it telescopes.

Step 4: Write a Partial Sum \(S_n\)

Express the partial sum \(S_n\) using the first \(n\) terms of the series:\[ S_n = (\ln 1 - \ln 2) + (\ln 2 - \ln 3) + \ldots + (\ln n - \ln(n+1)) \].

Step 5: Simplify the Partial Sum

Observe that most terms will cancel in this sequence, leaving:\[ S_n = \ln 1 - \ln(n+1) = -\ln(n+1) \].

Step 6: Analyze \(S_n\) as \(n\) Approaches Infinity

As \( n \to \infty \), \( \ln(n+1) \to \infty \), thus \( S_n = -\ln(n+1) \to -\infty \).

Step 7: Conclusion

Since \(S_n\) diverges to \(-\infty\), the series \( \sum_{k=1}^{\infty} \ln \frac{k}{k+1} \) diverges.

Key Concepts

Telescoping Series

A telescoping series is one where consecutive terms cancel each other out, leaving only a few terms when summed up. This is like a telescope collapsing into a compact form. This series can often reveal whether it converges or diverges by simplifying the expression first.

• To identify a telescoping series, look for terms that cancel with subsequent terms. For example, the series \( \sum_{k=1}^{n} (a_k - a_{k+1}) \).
• When expanded, many interior terms cancel out, simplifying to \(a_1 - a_{n+1}\).

In the given exercise, the series \( \sum_{k=1}^{\infty} \ln \frac{k}{k+1} \) is telescoping because \( \ln \frac{k}{k+1} \) is equivalent to \( \ln k - \ln(k+1) \).After canceling out the intermediate terms in a partial sum \( S_n \), we're left with \( \ln(1) - \ln(n+1) \), which simplifies the process of determining the series behavior.By writing its partial sum and analyzing it, we can determine that this series diverges, as its partial sum \( S_n \) approaches \(-\infty\) as \( n \) increases.

Harmonic Series

The harmonic series is a classic example in mathematics often used to illustrate concepts of divergence:\( \sum_{k=1}^{\infty} \frac{1}{k} \).Despite the terms getting smaller, their sum grows without bound as more terms are added.

• This series is significant because it provides a counterintuitive example of a diverging series that features terms approaching zero.
• The divergence of the harmonic series reminds us that not all series with diminishing terms converge.

Examining the harmonic series, you realize that even though the terms \( \frac{1}{k} \) get very small, their sum is not enough to turn the series into a convergent series.This concept can help understand why series similar to the one in the exercise, where terms become small, but follow a telescoping pattern, still diverge.

Logarithmic Function

Logarithmic functions are essential in numerous mathematical concepts. The natural logarithm function \( \ln(x) \) is defined for positive \( x \) and represents the power to which the base \( e \) (approximately 2.718) must be raised to get \( x \).

• A fundamental property of logarithms used in the exercise is: \( \ln \left( \frac{a}{b} \right) = \ln(a) - \ln(b) \).
• This property allows the rearrangement of terms in series to reveal latent telescoping behavior.

In the exercise, this property was crucial to express terms in a way that facilitated cancellation. The result was a much simpler form, which we analyzed to determine the behavior of the series.The logarithm's growth rate also influences how quickly the sum of series involving logarithms may diverge, as the logarithm increases at a slower rate than linear but still approaches infinity with increasing \( x \).