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Chapter 7: Q. 44 (page 653)

Use any convergence tests to determine whether the series converge absolutely, converge conditionally, or diverge. Explain why the series meets the hypotheses of the test you select.
${鈭�}_{k = 1}^{鈭�} {(- 1)}^{k} e^{- k}$

Short Answer

Expert verified

The series is absolutely convergent.

Step by step solution

01

Step 1. Given information.

Consider the given question,

${鈭�}_{k = 1}^{鈭�} {(- 1)}^{k} e^{- k}$

02

Step 2. Consider the ratio.

The general term of the series ${鈭�}_{k = 1}^{鈭�} a_{k} = {鈭�}_{k = 1}^{鈭�} {(- 1)}^{k} e^{- k}$ is $a_{k} = {(- 1)}^{k} e^{- k}$ .

The ratio $|\frac{a_{k + 1}}{a_{k}}|$ gives,

$|\frac{a_{k + 1}}{a_{k}}| = |\frac{{(- 1)}^{k + 1} e^{- (k + 1)}}{{(- 1)}^{k} e^{- k}}| |\frac{a_{k + 1}}{a_{k}}| = |\frac{(- 1) e^{- (k + 1)}}{e^{- k}}| |\frac{a_{k + 1}}{a_{k}}| = |\frac{1}{k}|$

03

Step 3. Find the value of the limit.

The value of $\lim_{k 鈫� 鈭�} |\frac{a_{k + 1}}{a_{k}}|$ is given below,

$\lim_{k 鈫� 鈭�} |\frac{a_{k + 1}}{a_{k}}| = \lim_{k 鈫� 鈭�} |\frac{1}{e}| = \frac{1}{e}$

The value of the above limit is less than $1$ .

Thus, by ratio test for absolute convergence, the series ${鈭�}_{k = 1}^{鈭�} {(- 1)}^{k} e^{- k}$ is absolutely convergent.

Hence, the series is absolutely convergent.

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Most popular questions from this chapter

Determine whether the series ${鈭�}_{k = 0}^{鈭�} \frac{2^{k + 2}}{5^{k - 1}}$ converges or diverges. Give the sum of the convergent series.

Let f(x) be a function that is continuous, positive, and decreasing on the interval $[1, 鈭�)$ such that role="math" localid="1649081384626" $\underset{x 鈫� 鈭�}{l i m} f (x) = 伪 > 0$ . What can the divergence test tell us about the series ${鈭�}_{k = 1}^{鈭�} f (k)$ ?

Let 0 < p < 1. Evaluate the limit $\lim_{k 鈫� 鈭�} \frac{1 / k \ln k}{1 / k^{p}}$

Explain why we cannot use a p-series with 0 < p < 1 in a limit comparison test to verify the divergence of the series ${鈭�}_{k = 2}^{鈭�} \frac{1}{k \log k}$

Improper Integrals: Determine whether the following improper integrals converge or diverge.

${鈭�}_{1}^{鈭�} \frac{1}{x} d x .$

Given that $a_{0} = - 3, a_{1} = 5, a_{2} = - 4, a_{3} = 2$ and ${鈭� a_{k}}_{k = 2}^{鈭�} = 7$ , find the value ofrole="math" localid="1648828803227" ${鈭� a_{k}}_{k = 3}^{鈭�}$ .

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