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Chapter 7: Q. 40 (page 657)

Conditional and absolute convergence: For each of the series that follow, determine whether the series converges absolutely, converges conditionally, or diverges. Explain the criteria you are using and why your conclusion is valid.
${鈭�}_{k = 1}^{鈭�} {(- 1)}^{k + 1} k$ ¹⁰⁰2^k

Short Answer

Expert verified

The series ${鈭�}_{k = 1}^{鈭�} {(- 1)}^{k + 1} \frac{k^{100}}{2^{k}}$ converges absolutely.

Step by step solution

01

Step 1. Given Information.

The series:

${鈭�}_{k = 1}^{鈭�} {(- 1)}^{k + 1} \frac{k^{100}}{2^{k}}$

02

Step 2. By Alternating Series Test.

According to the Alternating Series Test, the sequence $a_{k + 1} < a_{k}$ for every . Then the alternating series $a_{k + 1}, a_{k}$ both converges.

03

Step 3. Find ak+1.

$a_{k} = \frac{k^{100}}{2^{k}} a_{k + 1} = \frac{{(k + 1)}^{100}}{2^{k + 1}} a_{k + 1} < a_{k}$

So the sequence is monotonic decreasing sequence.

04

Step 4. Find limk→∞ak.

$\underset{k 鈫� 鈭�}{l i m} a_{k} = \underset{k 鈫� 鈭�}{l i m} \frac{k^{100}}{2^{k}} = 0$

So the series converges.

05

Step 5. Use Ratio test.

$\underset{k 鈫� 鈭�}{l i m} |\frac{a_{k + 1}}{a_{k}}| = \underset{k 鈫� 鈭�}{l i m} |{(- 1)}^{k + 1} (k + 1)|$ ¹⁰⁰2^k+1(-1)kk¹⁰⁰2^k=limk鈫�鈭�2k(k+1)100k1002k+1=limk鈫�鈭�(k+1)1002k100=12<1

By Ratio test, the series converges absolutely.

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Most popular questions from this chapter

Whenever a certain ball is dropped, it always rebounds to a height p% (0 < p < 100) of its original position. What is the total distance the ball travels before coming to rest when it is dropped from a height of h meters?

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