Chapter 7: Q. 47 (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.
$\sum_{k = 1}^{\infty} {(- 1)}^{k} e^{- k^{2}}$

Short Answer

Expert verified

The series converges absolutely.

Step by step solution

Step 1. Given information.

Consider the given question,

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

Step 2. Consider the general series.

The general term of the series $\sum_{k = 1}^{\infty} a_{k} = \sum_{k = 1}^{\infty} {(- 1)}^{k} e^{- k^{2}}$ is given by,

$a_{k} = {(- 1)}^{k} e^{- k^{2}}$

The ratio role="math" localid="1649154069574" $|\frac{a_{k + 1}}{a_{k}}|$ is given by,

role="math" localid="1649154169585" $|\frac{a_{k + 1}}{a_{k}}| = |\frac{{(- 1)}^{k + 1} e^{- {(k + 1)}^{2}}}{{(- 1)}^{k} e^{- k^{2}}}| = |\frac{(- 1) e^{- {(k + 1)}^{2}}}{e^{- k^{2}}}| = |\frac{e^{k^{2}}}{e^{{(k + 1)}^{2}}}|$

Step 3. Find the value of limit.

The value of $\lim_{k \to \infty} |\frac{a_{k + 1}}{a_{k}}|$ is given below,

role="math" localid="1649154467758" $\lim_{k \to \infty} |\frac{a_{k + 1}}{a_{k}}| = \lim_{k \to \infty} |\frac{e^{k^{2}}}{e^{{(k + 1)}^{2}}}| = \lim_{k \to \infty} |\frac{1}{e^{2 k + 1}}| = \lim_{k \to \infty} |\frac{1}{e \cdot e^{2 k}}| = 0$

The value of $\lim_{k \to \infty} |\frac{a_{k + 1}}{a_{k}}|$ is $0$ , which is less than $1$ .

Thus, by ratio test for absolute convergence, the series $\sum_{k = 1}^{\infty} {(- 1)}^{k} e^{- k^{2}}$ is absolutely convergent.

Hence, the given series is absolutely convergent.

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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.
$\sum_{k = 1}^{\infty} {(- 1)}^{k} e^{- k^{2}}$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Consider the general series.

Step 3. Find the value of limit.

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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.∑k=1∞-1ke-k2

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Consider the general series.

Step 3. Find the value of limit.

One App. One Place for Learning.

Most popular questions from this chapter

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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.
$\sum_{k = 1}^{\infty} {(- 1)}^{k} e^{- k^{2}}$