Chapter 7: Q. 25 (page 653)

Use the alternating series test to determine whether the series in Exercises 24–29 converge or diverge. If a series converges, determine whether it converges absolutely or conditionally.
$\sum_{k = 0}^{\infty} \frac{(- 1)^{k + 1}}{1 + k^{2}}$

Short Answer

Expert verified

The series $\sum_{k = 0}^{\infty} \frac{(- 1)^{k + 1}}{1 + k^{2}}$ converges

Step by step solution

Step 1. Given information

$\sum_{k = 0}^{\infty} \frac{(- 1)^{k + 1}}{1 + k^{2}}$

Step 2. Solving the series

$\sum_{k = 0}^{\infty} \frac{(- 1)^{k + 1}}{1 + k^{2}} = \sum_{k = 1}^{\infty} (- 1)^{k + 1} a_{k} where a_{k} = \frac{1}{1 + k^{2}}$

Substitute $k = k + 1$ in $a_{k} = \frac{1}{1 + k^{2}}$ ,

$a_{k + 1} = \frac{(k + 1)^{k + 1}}{(k + 1)!}$

Step 3. Calculate the limit

$\begin{aligned} lim_{k \to \infty} a_{k} = lim_{k \to \infty} \frac{1}{1 + k^{2}} \\ = 0 \end{aligned}$

Therefore, $\sum_{k = 0}^{\infty} \frac{(- 1)^{k + 1}}{1 + k^{2}}$ converges absolutely.

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Use the alternating series test to determine whether the series in Exercises 24–29 converge or diverge. If a series converges, determine whether it converges absolutely or conditionally.
$\sum_{k = 0}^{\infty} \frac{(- 1)^{k + 1}}{1 + k^{2}}$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Solving the series

Step 3. Calculate the limit

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Use the alternating series test to determine whether the series in Exercises 24–29 converge or diverge. If a series converges, determine whether it converges absolutely or conditionally.∑k=0∞ (−1)k+11+k2

Short Answer

Step by step solution

Step 1. Given information

Step 2. Solving the series

Step 3. Calculate the limit

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

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Study anywhere. Anytime. Across all devices.

Use the alternating series test to determine whether the series in Exercises 24–29 converge or diverge. If a series converges, determine whether it converges absolutely or conditionally.
$\sum_{k = 0}^{\infty} \frac{(- 1)^{k + 1}}{1 + k^{2}}$