Q. 9 Complete Example 4 by showing th... [FREE SOLUTION]

Chapter 8: Q. 9 (page 669)

Complete Example 4 by showing that the power series ${鈭�}_{k = 0}^{鈭�} 鈥� (鈭� 1)^{k} \frac{k}{3 k + 1} (2 x 鈭� 3)^{k}$ diverges when $x = 2$ .

Short Answer

Expert verified

Ans: By the divergence test, the power series ${鈭�}_{k = 0}^{鈭�} 鈥� (鈭� 1)^{k} \frac{k}{3 k + 1} (2 x 鈭� 3)^{k}$ diverges when $x = 2$

Step by step solution

Step 1. Given information.

given,

${鈭�}_{k = 0}^{鈭�} 鈥� (鈭� 1)^{k} \frac{k}{3 k + 1} (2 x 鈭� 3)^{k}$

Step 2. We evaluate the series when x=2

So,

$\begin{aligned} {鈭�}_{k = 0}^{鈭�} 鈥� (鈭� 1)^{k} \frac{k}{3 k + 1} (2 x 鈭� 3)^{k} = {鈭�}_{k = 0}^{鈭�} 鈥� (鈭� 1)^{k} \frac{k}{3 k + 1} [2 (2) 鈭� 3]^{k} \\ = {鈭�}_{k = 0}^{鈭�} 鈥� (鈭� 1)^{k} \frac{k}{3 k + 1} \end{aligned}$

Therefore, $k 鈫� 鈭�$ , ${鈭�}_{k = 0}^{鈭�} 鈥� (鈭� 1)^{k} \frac{k}{3 k + 1} 鈫� 鈭�$

Step 3. Thus,

Thus, by the divergence test, the power series ${鈭�}_{k = 0}^{鈭�} 鈥� (鈭� 1)^{k} \frac{k}{3 k + 1} (2 x 鈭� 3)^{k}$ diverges when $x = 2$

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Complete Example 4 by showing that the power series ${鈭�}_{k = 0}^{鈭�} 鈥� (鈭� 1)^{k} \frac{k}{3 k + 1} (2 x 鈭� 3)^{k}$ diverges when $x = 2$ .

Short Answer

Step by step solution

Step 1. Given information.

Step 2. We evaluate the series when x=2

Step 3. Thus,

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91影视

Complete Example 4 by showing that the power series 鈭�k=0鈭�鈥�(鈭�1)kk3k+1(2x鈭�3)kdiverges when x=2.

Short Answer

Step by step solution

Step 1. Given information.

Step 2. We evaluate the series when x=2

Step 3. Thus,

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Statistics

Decision Maths

Mechanics Maths

Geometry

Calculus

Study anywhere. Anytime. Across all devices.

Complete Example 4 by showing that the power series ${鈭�}_{k = 0}^{鈭�} 鈥� (鈭� 1)^{k} \frac{k}{3 k + 1} (2 x 鈭� 3)^{k}$ diverges when $x = 2$ .