Problem 11 - Thomas Calculus

Question

In Exercises \(1 - 14 ,\) determine if the alternating series converges or diverges. Some of the series do not satisfy the conditions of the Alternating Series Test. $$ \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { \ln n } { n } $$

Step-by-Step Solution

Answer

The series converges by the Alternating Series Test.

Step 1: Define the Alternating Series

The given series is \( \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { \ln n } { n } \). This is an alternating series because it changes sign with each term due to the \( ( - 1 ) ^ { n + 1 } \) factor. We denote the terms as \( a_n = \frac{\ln n}{n} \).

Step 2: Check Decreasing Nature of Terms

For the Alternating Series Test to apply, \( a_n \) should be decreasing for all sufficiently large \( n \). We check if \( \frac{\ln n}{n} \) is decreasing by finding its derivative and checking its sign. The derivative \( f'(n) = \frac{1 - \ln n}{n^2} \). For \( n \geq 3 \), \( 1 - \ln n \le 0 \), so \( f'(n) \le 0 \). Thus, \( \frac{\ln n}{n} \) is a decreasing function for \( n \geq 3 \).

Step 3: Check Limit Condition

According to the Alternating Series Test, \( \lim_{{n \to \infty}} a_n = 0 \) must hold. Compute \( \lim_{{n \to \infty}} \frac{\ln n}{n} \). As \( n \to \infty \), \( \ln n \to \infty \), but it grows much slower than \( n \). Applying L'Hôpital's rule, the limit is \( \lim_{n \to \infty} \frac{1/n}{1} = 0 \). Thus, the condition is satisfied.

Step 4: Conclude with the Alternating Series Test

Since both conditions for the Alternating Series Test are met (\( a_n \) is decreasing for \( n \ge 3 \) and \( \lim_{{n \to \infty}} a_n = 0 \)), the series \( \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { \ln n } { n } \) converges.

Key Concepts

Convergence

Convergence is a fundamental concept in calculus and series analysis.
It refers to the behavior of a series as the number of terms approaches infinity. Specifically, a series converges if the sum of its terms approaches a finite limit.
In the case of an alternating series, convergence can be determined using the Alternating Series Test. This test requires

• The sequence of terms to be decreasing,
• And the terms must approach zero as they extend to infinity.

If both conditions are met, the series converges.
In the given series, \(\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { \ln n } { n }\), both conditions are satisfied, leading us to conclude that the series converges.

Series

A series is the sum of the terms of a sequence.
Different from a sequence, which is a list of numbers, a series adds up elements to form a sum.
The series we're discussing is an alternating series, specifically denoted as \(\sum _ { n = 1 } ^ { \infty } ( - 1 )^{n+1} \frac { \ln n } { n } \).
Alternating series change form from positive to negative due to the \(( - 1 )^{n+1}\) component.
Such a sign change alerts us to potentially apply the Alternating Series Test. This test helps in determining whether this back-and-forth pattern eventually yields a convergent result.

L'Hôpital's Rule

L'Hôpital's Rule is a valuable technique in calculus for finding limits of indeterminate forms.
Typically used when both the numerator and denominator of a fraction approach zero or infinity.
It involves differentiating the numerator and the denominator separately.
In our exercise, to find \(\lim_{{n \to \infty}} \frac{\ln n}{n}\), we used L'Hôpital's Rule because both the numerator \(\ln n\) and denominator \(n\) approach infinity.
After applying the rule

• Differentiate \(\ln n\) to get \(1/n\),
• And \(n\) to get \(1\).

The result simplifies to \(\lim_{n \to \infty} \frac{1/n}{1} = 0\).
This confirms the limit condition required in the Alternating Series Test.

Derivatives

Derivatives represent the rate of change of a function with respect to a variable.
It is a core concept in calculus, providing insights into the behavior of functions.
In determining if \( \frac{\ln n}{n} \) is decreasing, we calculate its derivative.
The derivative, \( f'(n) = \frac{1 - \ln n}{n^2} \), shows us the slope of \(\frac{\ln n}{n}\).
Evaluating this derivative helps establish whether the function is increasing or decreasing.
In our series, we found that for \( n \ge 3\), \(1 - \ln n\) is negative, making \( f'(n)\) negative.
This negativity indicates a decreasing function, thereby satisfying one of the conditions for the Alternating Series Test.