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) Diverges. (b) Converges. (c) Converges. (d) Converges. (e) Converges. (f) Converges.

Step 1: Evaluate Series (a)

Consider the series \( \sum_{n=1}^{\infty} \ln \left(1+\frac{1}{n}\right) \). We can use a property of logarithms: \( \ln(1+x) \approx x \) for small \( x \). Thus, \( \ln \left(1+\frac{1}{n}\right) \approx \frac{1}{n} \). The series \( \sum \frac{1}{n} \) is the harmonic series, which is known to diverge. Hence, series (a) diverges.

Step 2: Simplify and Evaluate Series (b)

For the series \( \sum_{n=1}^{\infty} \ln \left[\frac{(n+1)^{2}}{n(n+2)}\right] \), simplify the logarithm: \[ \ln \left[\frac{(n+1)^{2}}{n(n+2)}\right] = \ln(n+1)^2 - \ln(n) - \ln(n+2). \] This becomes \( 2\ln(n+1) - \ln(n) - \ln(n+2) \). Notice this is a telescoping series: terms cancel out successively. Hence, the series converges to a constant value.

Step 3: Evaluate Series (c)

The series \( \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{\ln n}} \) is examined for convergence. Observe that for large \( n \), \( (\ln n)^{\ln n} \) grows very rapidly, more than exponentially. Compare with a p-series: if \( p > 1 \), \( \sum \frac{1}{n^p} \) converges. Here, because the denominator grows faster than any polynomial rate, the series converges.

Step 4: Evaluate Series (d)

For the series \( \sum_{n=3}^{\infty} \frac{1}{[\ln (\ln n)]^{\ln n}} \), \( [\ln (\ln n)]^{\ln n} \) also grows extremely rapidly as \( n \to \infty \). Similar to the logic in series (c), as this grows more than exponentially, the series converges.

Step 5: Evaluate Series (e)

The series \( \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{4}} \) corresponds to a p-series test with terms in the form \( \frac{1}{n^{p}}, p=4 \) for logarithmic terms. Since \( p = 4 > 1 \), by the p-series test, the series converges.

Step 6: Evaluate Series (f)

Consider the series \( \sum_{n=1}^{\infty} \left[\frac{\ln n}{n}\right]^{2} = \sum_{n=1}^{\infty} \frac{(\ln n)^{2}}{n^{2}} \). We can apply the comparison test: compare with \( \sum \frac{1}{n^{3/2}} \), which converges. Since \( \ln n\) grows slower than any polynomial rate, the series converges.

Key Concepts

Harmonic Series

The harmonic series is one of the most important series in mathematics. It takes the form \( \\sum_{n=1}^{\infty} \frac{1}{n} \), where each term in the series is the reciprocal of a positive integer.
This series is noteworthy because it diverges, meaning that as you keep adding terms, the sum grows indefinitely.
You might expect it to converge, given that the terms get very small, but their cumulative effect still leads to infinity.
This divergence can be understood by approximating the series as an integral, which also diverges.

Telescoping Series

Telescoping series are interesting because many terms cancel out, simplifying the series significantly. Consider a series where the terms are structured such that components of different terms cancel each other out.
For example: \[ \sum_{n=1}^{\infty} \ln \left( \frac{(n+1)^2}{n(n+2)} \right) \] can be rewritten as \( 2\ln(n+1) - \ln(n) - \ln(n+2) \).
With telescoping series, one usually ends up with just a few terms remaining, which makes it easy to find the limit or sum of the series. These characteristics are what make telescoping series a powerful tool when analyzing series for convergence.

P-Series Test

The p-series test is a straightforward method used to test the convergence of series that have terms of the form \( \\sum_{n=1}^{\infty} \frac{1}{n^p} \). It's very efficient because it allows you to determine convergence based solely on the exponent \( p \).
The rule is simple: if \( p > 1 \), the series converges. If \( p \leq 1 \), the series diverges.
This test is useful for quickly assessing many series, including variations where the term \( n^p \) is replaced by other functions that grow or decrease rapidly. For example, adjusting the series to \( \\sum_{n=2}^{\infty} \frac{1}{(\ln n)^4} \) entails using the principle of comparison to assess it against a standard p-series.

Comparison Test

The comparison test is a handy tool to determine the convergence or divergence of series by comparing them to another series that is already known to converge or diverge.

• If a series \( a_n \) is less than or equal to a known convergent series \( b_n \) from some point onward, then \( a_n \) must also converge.
• Conversely, if \( a_n \) is greater than or equal to a known divergent series \( b_n \) from some point onward, then \( a_n \) must also diverge.

In practice, you often use the comparison test on series where the terms look complicated but can be bounded by a simpler series.