Problem 34 - Calculus: Early Transcendentals

Question

For what value of \( p \) is each series convergent? \( \displaystyle \sum_{n = 2}^{\infty} ( - 1)^{n-1} \frac {(\ln n)^P}{n} \)

Step-by-Step Solution

Answer

The series converges for \( P > 0 \).

Step 1: Identify the Series Type

The given series \( \sum_{n=2}^{\infty} (-1)^{n-1} \frac{(\ln n)^P}{n} \) is an alternating series. It alternates in sign because of the \((-1)^{n-1}\) factor.

Step 2: Apply the Alternating Series Test

For an alternating series \( \sum (-1)^{n} a_n \) to converge, two conditions must be met: 1) \( a_n > a_{n+1} \) for all \( n \) eventually, and 2) \( \lim_{n \to \infty} a_n = 0 \). For our series, \( a_n = \frac{(\ln n)^P}{n} \).

Step 3: Determine if the Terms Tend to Zero

Evaluate \( \lim_{n \to \infty} \frac{(\ln n)^P}{n} \). If \( P > 0 \), the numerator grows slower than the denominator, making \( \lim_{n \to \infty} \frac{(\ln n)^P}{n} = 0 \). Thus, for \( P > 0 \), the terms tend to zero.

Step 4: Check Monotonicity of \( a_n \)

To check that \( \frac{(\ln n)^P}{n} \) is decreasing, check the derivative of \( f(n) = \frac{(\ln n)^P}{n} \) for \( n \geq 2 \). Use the quotient rule to differentiate. The result will show that \( f(n) \) is eventually decreasing for \( P > 0 \).

Step 5: Conclusion

Since the series satisfies both conditions of the Alternating Series Test for \( P > 0 \) (the sequence tends to zero and is eventually decreasing), the series converges for any \( P > 0 \).

Key Concepts

Convergence of Series

Series convergence is a fundamental concept in calculus, dealing with whether the sum of an infinite list of numbers reaches a finite limit. For the series to converge, the partial sums need to approach a specific value as the number of terms goes to infinity.
In the given problem, the series is written as:

• \( \displaystyle \sum_{n=2}^{\infty} (-1)^{n-1} \frac{(\ln n)^P}{n} \)

The task is to find for which values of \( P \) the series converges.
One important method of checking for convergence is the use of tests, such as the Alternating Series Test, which assesses specific conditions for convergence based on the behavior of its terms.

Differentiation using Quotient Rule

Differentiating using the quotient rule is a technique used when you have a function that is the ratio of two differentiable functions. The quotient rule is written as:

• If \( f(x) = \frac{u(x)}{v(x)} \), then \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \)

In our case, we need to differentiate

• \( f(n) = \frac{(\ln n)^P}{n} \), where \( u(n) = (\ln n)^P \) and \( v(n) = n \).

Finding the derivatives:

• \( u'(n) = P(\ln n)^{P-1} \cdot \frac{1}{n} \)
• \( v'(n) = 1 \)

Using the quotient rule:

• \( f'(n) = \frac{\left(P(\ln n)^{P-1} \cdot \frac{1}{n}\right) \cdot n - (\ln n)^P \cdot 1}{n^2} \)

Thus, this differentiation helps in determining if the function is monotonically decreasing, a requirement for the Alternating Series Test.

Alternating Series

An alternating series is a series whose terms alternate in sign; it often involves sequences like

• \( (-1)^n \) or \( (-1)^{n-1} \).

The Alternating Series Test (or Leibniz's test) provides a way to determine convergence for alternating series. The conditions are:

• The absolute value of the terms \( a_n \) must decrease monotonically.
• The limit of the absolute value of the terms as \( n \to \infty \) must be zero, i.e., \( \lim_{n \to \infty} a_n = 0 \).

Applying the test to our series, \( \sum_{n=2}^{\infty} (-1)^{n-1} \frac{(\ln n)^P}{n} \), the conditions confirm that:

• The terms tend to zero for \( P > 0 \).
  Thus, the series converges if \( P > 0 \), given that the sequence of absolute values decreases as determined using differentiation techniques such as the quotient rule.