Problem 26 - Essential Calculus Early Transcendentals

Question

Determine whether the series is convergent or divergent by expressing \(s_{n}\) as a telescoping sum (as in Example 6\() .\) If it is convergent, find its sum. $$\sum_{n=1}^{\infty} \ln \frac{n}{n+1}$$

Step-by-Step Solution

Answer

The series is divergent because its sum approaches \(-\infty\).

Step 1: Write the Series as a Telescoping Sum

The given series is \( \sum_{n=1}^{\infty} \ln \frac{n}{n+1} \). We can rewrite each term **\(\ln \frac{n}{n+1}\) ** using the properties of logarithms: \( \ln \frac{n}{n+1} = \ln n - \ln(n+1) \). Thus, the series becomes \( \sum_{n=1}^{\infty} (\ln n - \ln(n+1)) \).

Step 2: Identify Terms That Cancel

Write out the first few terms of the series: \( (\ln 1 - \ln 2) + (\ln 2 - \ln 3) + (\ln 3 - \ln 4) + \cdots \). Notice that in this telescoping series, each term \(\ln n \) cancels with the \( -\ln n \) from the next term.

Step 3: Analyze the Remaining Terms

Since the series is telescoping, most intermediate terms cancel out, leaving only the first term of the first log and the negative of the last term from the last expanded log term. For the partial sum \(s_N\), this simplifies to \( \ln 1 - \ln(N+1) \).

Step 4: Determine the Convergence of the Series

Evaluate the limit of the partial sum as \( N \to \infty \): \[ s_N = \ln 1 - \ln(N+1) = 0 - \ln(N+1). \]As \( N \to \infty \), \( \ln(N+1) \to \infty \), thus \( s_N \to -\infty \).

Step 5: Conclusion on Convergence

Because \( s_N \to -\infty \), the series diverges. A convergent series will have a finite sum, but this series does not meet that criterion.

Key Concepts

Convergent Series

Understanding convergent series is crucial for identifying whether a series sums to a specific finite value. A series is considered convergent if the sum of its infinite terms approaches a determined number, meaning it does not continue to grow indefinitely.
One of the primary methods to test for convergence is the telescoping nature of the series. This allows for partial sums to cancel out intermediate terms, simplifying the analysis. In a convergent telescoping series, the limited terms left after cancellation should sum up to a finite value.

• Analyzing the remainder terms post-cancellation is essential.
• Ensure that as more terms are added, the sum settles to a particular value.
• If the sum goes to infinity, it indicates divergence rather than convergence.

For example, if the terms consolidate to a cancellation loop producing a finite result, you've got convergence. The sum stabilizes with each added term, confirming its finite nature. However, if as in the given problem the remaining terms diverge, reaching infinity, your series is not convergent.

Divergent Series

Divergent series behave oppositely to convergent series. Here, the sequence of partial sums does not settle at any finite value. Instead, it grows without bound, often resulting in infinite sums.
Telescoping series can sometimes simplify analysis by canceling intermediate terms. However, the remaining terms tell the real story of a series' divergence.

• If after removing intermediate terms, a valid remaining computation heads towards infinity, it suggests divergence.
• Unbounded series values vastly surpass finite limits present in convergent series.
• In typical divergent series, partial sums continue to rise, with no horizontal asymptote to stabilize the sum.

In the given example, the telescoping provided a remainder of \( \ln(N+1) \). This suggests the series will diverge to \( -\infty \) because the natural logarithm of an indefinitely increasing number also goes towards infinity.

Logarithmic Properties

Logarithms transform products into sums and quotients into differences. This essential property is utilized especially in complex mathematical topics like telescoping series.
The main logarithmic property applied in the series under discussion is: \[ \ln\left(\frac{a}{b}\right) = \ln a - \ln b \]. This transformation enabled us to express original series terms differently, forming a telescoping sequence.

• Breaking down a logarithm of a fraction simplifies the log term into easier, computable units.
• Understanding how to manipulate logarithmic expressions through properties can unravel complex series.
• Mastery of these transformations is pivotal in simplifying terms within series.

In particular for your exercise, applying \( \ln n = \ln(n+1) \) directly helped to set up a telescoping form, allowing cancellation, thus simplifying the convergence/divergence analysis. Remember, properties of logs bridge the gap between complex algebraic expressions and simpler, approachable math problems.