Q. 10 Show that the power series 鈭慿=... [FREE SOLUTION]

Chapter 8: Q. 10 (page 669)

Show that the power series ${鈭�}_{k = 0}^{鈭�} 鈥� \frac{(鈭� 1)^{k}}{2 k + 1} x^{2 k + 1}$ converges conditionally when $x = 1$ and when $x = - 1$ . What does this behavior tell you about the interval of convergence for the series?

Short Answer

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Ans: The power series ${鈭�}_{k = 0}^{鈭�} 鈥� \frac{(鈭� 1)^{k}}{2 k + 1} x^{2 k + 1}$ has the interval of convergence $[- 1, 1]$

Step by step solution

Step 1. Given information.

given,

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

Step 2. Evaluate the series when x=1

so,

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

So, for $x = 1$ , we have the alternating harmonic series which converges conditionally.

Step 3. Evaluate the series when x=-1

So,

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

So, for localid="1649356608759" $x = - 1$ , we have the alternating harmonic series which converges conditionally.

Step 4. Thus,

Therefore, the power series ${鈭�}_{k = 0}^{鈭�} 鈥� \frac{(鈭� 1)^{k}}{2 k + 1} x^{2 k + 1}$ has the interval of convergence $[- 1, 1]$ .

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91影视

Show that the power series ${鈭�}_{k = 0}^{鈭�} 鈥� \frac{(鈭� 1)^{k}}{2 k + 1} x^{2 k + 1}$ converges conditionally when $x = 1$ and when $x = - 1$ . What does this behavior tell you about the interval of convergence for the series?

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Evaluate the series when x=1

Step 3. Evaluate the series when x=-1

Step 4. Thus,

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91影视

Show that the power series 鈭�k=0鈭�鈥�(鈭�1)k2k+1x2k+1converges conditionally when x=1and when x=-1. What does this behavior tell you about the interval of convergence for the series?

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Evaluate the series when x=1

Step 3. Evaluate the series when x=-1

Step 4. Thus,

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Calculus

Mechanics Maths

Applied Mathematics

Pure Maths

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Show that the power series ${鈭�}_{k = 0}^{鈭�} 鈥� \frac{(鈭� 1)^{k}}{2 k + 1} x^{2 k + 1}$ converges conditionally when $x = 1$ and when $x = - 1$ . What does this behavior tell you about the interval of convergence for the series?