Chapter 8: Q. 12 (page 669)

Show that the power series $\sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k^{2}} x^{k}$ converges absolutely when $x = 1$ and when $x = - 1$ . What does this behavior tell you about the interval of convergence for the series?

Short Answer

Expert verified

Ans: The power series $\sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k^{2}} x^{k}$ has the interval of convergence $[- 1, 1]$

Step by step solution

Step 1. Given information.

given,

$\sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k^{2}} x^{k}$

Step 2. Evaluate the series when x=1

So,

$\begin{aligned} \sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k^{2}} x^{k} = \sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k^{2}} (1)^{k} \\ = \sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k^{2}} \end{aligned}$

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

Step 3. We evaluate the series when x=-1

So,

$\begin{aligned} \sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k^{2}} x^{k} = \sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k^{2}} (- 1)^{k} \\ = \sum_{k = 1}^{\infty} \frac{(- 1)^{2 k}}{k^{2}} \\ = \sum_{k = 1}^{\infty} \frac{1}{k^{2}} \end{aligned}$

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

Step 4. Thus,

Therefore, the power series $\sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k^{2}} x^{k}$ has the interval of convergence $[- 1, 1]$ .

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Show that the power series $\sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k^{2}} x^{k}$ converges absolutely 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. We evaluate the series when x=-1

Step 4. Thus,

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Show that the power series ∑k=1∞ (−1)kk2xkconverges absolutely 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. We evaluate the series when x=-1

Step 4. Thus,

One App. One Place for Learning.

Most popular questions from this chapter

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Show that the power series $\sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{k^{2}} x^{k}$ converges absolutely when $x = 1$ and when $x = - 1$ . What does this behavior tell you about the interval of convergence for the series?