(In n)ªProblem 2. Use the Alternating Series Test to show that for any real number a, the series >(-–1)"+1nn=1converges. (Hint: You need to show that the sequence {(ln n)ª/n} converges to 0 for any a, and is eventuallydecreasing. Show that the function f(x) = (In x)ª /x is decreasing if x is sufficiently large.)

The given series is : ∑n=1∞-1n+1ln nan Applying Alternative series test to the above series is…

Answered: (In n)ª Problem 2. Use the Alternating…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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(In n)ª
Problem 2. Use the Alternating Series Test to show that for any real number a, the series >(-–1)"+1
n
n=1
converges. (Hint: You need to show that the sequence {(ln n)ª/n} converges to 0 for any a, and is eventually
decreasing. Show that the function f(x) = (In x)ª /x is decreasing if x is sufficiently large.)

Expert Solution

Step 1

The given series is :

$\sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \frac{{(\ln n)}^{a}}{n}$

Applying Alternative series test to the above series is given as:

The first condition is given as:

$\underset{n \to \infty}{l i m} b_{n} = 0$

Here, $b_{n} = \frac{{(\ln n)}^{a}}{n}$

$\begin{array}{rcl} \underset{n \to \infty}{l i m} b_{n} & = & \underset{n \to \infty}{l i m} \frac{{(\ln n)}^{a}}{n} \\ = & \underset{n \to \infty}{l i m} \frac{a (\ln n)}{n} \\ = & \underset{n \to \infty}{l i m} \frac{\frac{a}{x}}{1} \\ = & 0 \end{array}$

This satisfies the first condition.

The second condition is given as :

$b_{n}$ must be a decreasing sequence.

$\begin{array}{rcl} b_{n} & = & f (n) \\ f' (n) & = & \frac{n \times \frac{1}{n} \times a \ln n - {(\ln n)}^{a}}{n^{2}} \\ = & \frac{a \ln n - {(\ln n)}^{a}}{n^{2}} \end{array}$

Thus, $f' (n) \leq 0$

So, $f$ is a decreasing sequence so $f (n + 1) < f (n)$

Hence, $b_{n + 1} < b_{n}$

This satisfies the second condition.

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