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Chapter 7: Q. 16 (page 624)

Use the divergence test to analyze the given series. Each answer should either be the series diverges or the divergence test fails, along with the reason for your answer.
${鈭�}_{k = 1}^{鈭�} \frac{1}{k}$

Short Answer

Expert verified

The divergence test failed as $\lim_{k 鈫� 鈭�} \frac{1}{k} = 0$ .

Step by step solution

01

Step 1. Given information.

Consider the given series,

${鈭�}_{k = 1}^{鈭�} \frac{1}{k}$

02

Step 2. Analyze the series.

The divergence test states that if the sequence $\{a_{k}\}$ does not converge to zero, then the series ${鈭� a_{k}}_{k = 1}^{鈭�}$ diverges.

The value of the sequence $\{a_{k}\} = \{\frac{1}{k}\}$ is given below,

$\lim_{k 鈫� 鈭�} a_{k} = \lim_{k 鈫� 鈭�} \frac{1}{k} = 0$

Thus, the divergence test failed as $\lim_{k 鈫� 鈭�} \frac{1}{k} = 0$ .

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Most popular questions from this chapter

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" ${鈭� a_{k}}_{k = 3}^{鈭�}$ .

For a convergent series satisfying the conditions of the integral test, why is every remainder $R_{n}$ positive? How can $R_{n}$ be used along with the term $S_{n}$ from the sequence of partial sums to understand the quality of the approximation $S_{n}$ ?

Prove Theorem 7.25. That is, show that the series ${鈭�}_{k = 1}^{鈭�} a_{k} a n d {鈭�}_{k = M}^{鈭�} a_{k}$ either both converge or both diverge. In addition, show that if ${鈭�}_{k = M}^{鈭�} a_{k}$ converges to L, then ${鈭�}_{k = 1}^{鈭�} a_{k}$ converges tolocalid="1652718360109" $a_{1} + a_{2} + a_{3} + . . . . + a_{M - 1} + L .$

Use the divergence test to analyze the given series. Each answer should either be the series diverges or the divergence test fails, along with the reason for your answer.

${鈭�}_{k = 1}^{鈭�} 鈥� \frac{k^{2} + 1}{k!}$

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.

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