Problem 43 - Calculus Early Transcendentals: Pearson New International Edition

Question

Test for convergence or divergence. In some cases, a clever manipulation using the properties of logarithms will simplify the problem. (a) \(\sum_{n=1}^{\infty} \ln \left(1+\frac{1}{n}\right)\) (b) \(\sum_{n=1}^{\infty} \ln \left[\frac{(n+1)^{2}}{n(n+2)}\right]\) (c) \(\sum_{n=2}^{\infty} \frac{1}{(\ln n)^{\ln n}}\) (d) \(\sum_{n=3}^{\infty} \frac{1}{[\ln (\ln n)]^{\ln n}}\) (e) \(\sum_{n=2}^{\infty} \frac{1}{(\ln n)^{4}}\) (f) \(\sum_{n=1}^{\infty}\left[\frac{\ln n}{n}\right]^{2}\)

Step-by-Step Solution

Answer

(a) Divergent, (b) Convergent, (c) Convergent, (d) Convergent, (e) Convergent, (f) Convergent.

Step 1: Analyze Part (a)

Consider the series \(\sum_{n=1}^{\infty} \ln \left(1+\frac{1}{n}\right)\). Notice that \(\ln \left(1+\frac{1}{n}\right) \approx \frac{1}{n}\) for large \(n\). Hence, the series \(\sum_{n=1}^{\infty} \ln \left(1+\frac{1}{n}\right)\) behaves like a harmonic series, which is divergent.

Step 2: Analyze Part (b)

Consider the series \(\sum_{n=1}^{\infty} \ln \left[\frac{(n+1)^{2}}{n(n+2)}\right]\). Simplify the expression: \(\ln \left[\frac{(n+1)^{2}}{n(n+2)}\right] = 2\ln(n+1) - \ln(n) - \ln(n+2)\). Observe the telescoping nature of this series. The series converges as each term cancels out previous terms.

Step 3: Analyze Part (c)

Consider the series \(\sum_{n=2}^{\infty} \frac{1}{(\ln n)^{\ln n}}\). For large \(n\), \((\ln n)^{\ln n}\) grows extremely fast, which makes \(\frac{1}{(\ln n)^{\ln n}}\) decrease very rapidly. Therefore, the series converges.

Step 4: Analyze Part (d)

Consider the series \(\sum_{n=3}^{\infty} \frac{1}{[\ln (\ln n)]^{\ln n}}\). Like in part (c), the denominator grows very rapidly, leading to each term becoming very small. Thus, the series converges.

Step 5: Analyze Part (e)

Consider the series \(\sum_{n=2}^{\infty} \frac{1}{(\ln n)^{4}}\). Use the integral test. The function \(\frac{1}{(\ln x)^{4}}\) is continuous, positive, and decreasing for \(x \geq 2\). The integral \(\int_{2}^{\infty} \frac{1}{(\ln x)^{4}} \, dx\) converges, so the series converges.

Step 6: Analyze Part (f)

Consider the series \(\sum_{n=1}^{\infty}\left[\frac{\ln n}{n}\right]^{2}\). Note that for large \(n\), this behaves like the series \(\sum \frac{(\ln n)^{2}}{n^{2}}\). Compare it with \(\sum \frac{1}{n^{2}}\), a convergent p-series. Since \(\sum \frac{(\ln n)^2}{n^2}\) converges, the original series also converges.

Key Concepts

Logarithms Properties

Understanding the properties of logarithms can greatly simplify the evaluation of convergence in sequences and series. Here, we often make use of the properties:

• \( \ln(ab) = \ln a + \ln b \)
• \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \)
• \( \ln(a^n) = n\ln a \)

These properties are particularly valuable when decomposing complex logarithmic expressions or when they are part of a fraction. For example, when evaluating series like \( \sum_{n=1}^{\infty} \ln \left( 1 + \frac{1}{n} \right) \), recognizing that \( \ln(1 + \frac{1}{n}) \approx \frac{1}{n} \) for large \( n \) shows it behaves like the harmonic series, aiding in determining divergence.

Harmonic Series

The harmonic series is an infinite series expressed as \[ \sum_{n=1}^{\infty} \frac{1}{n} \].Despite its simplicity, the harmonic series is an important example as it diverges. While individual terms \( \frac{1}{n} \) approach zero, the series itself extends indefinitely without bound.
A common property utilized is the comparison test, where the series is compared to another series. In the case where a series matches the behavior of the harmonic series, like in \( \sum_{n=1}^{\infty} \ln \left(1+\frac{1}{n}\right) \), we identify the similar divergent behavior.

Telescoping Series

A telescoping series is one where consecutive terms cancel each other out, leaving a simpler expression to evaluate. Our goal is to rearrange or simplify the terms so that almost every term cancels with another.
The series \( \sum_{n=1}^{\infty} \ln \left[ \frac{(n+1)^2}{n(n+2)} \right] \) simplifies to \( 2\ln(n+1) - \ln(n) - \ln(n+2) \).When expressed in this form, you'll notice a pattern: each \( -\ln(n) \) cancels with \( \ln(n) \) from the next term. Such cancellations simplify the analysis and often lead to identifying convergence.

Integral Test

The integral test is a method to determine the convergence of infinite series by comparing it to a related integral. If \( f(x) \) is a continuous, positive, decreasing function for \( x \geq 1 \) and \[ a_n = f(n), \]then the series \[ \sum_{n=1}^{\infty} a_n \]converges if and only if the integral \[ \int_{1}^{\infty} f(x) \, dx \]converges.
This method is beneficial because integrals can sometimes be easier to evaluate than the series itself. For instance, in evaluating \( \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{4}} \), the integral \( \int_{2}^{\infty} \frac{1}{(\ln x)^{4}} \, dx \) converges, indicating that the series does too.