Q. 13 Let 0 < p < 1. Evaluate th... [FREE SOLUTION]

Chapter 7: Q. 13 (page 631)

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 ${鈭�}_{k = 2}^{鈭�} \frac{1}{k \log k}$

Short Answer

Expert verified

$The p - series is divergent for 0 < p < 1 . The limit comparison test if the ratio is zero states that if \lim_{k 鈫� 鈭�} \frac{a_{k}}{b_{k}} and {鈭�}_{k = 1}^{鈭�} b_{k} converges, then {鈭�}_{k = 1}^{鈭�} a_{k} converges The limit comparison test fails to give any information about the divergence of the series {鈭�}_{k = 2}^{鈭�} \frac{1}{k lnk}$

Step by step solution

Step 1. Given

${鈭�}_{k = 2}^{鈭�} \frac{1}{k \log k}$

Step 2.Finding the value of the expression

$The value of the expression is \lim_{k 鈫� 鈭�} \frac{1 / k lnk}{1 / k^{p}} \lim_{k 鈫� 鈭�} \frac{1 / k lnk}{1 / k^{p}} = \lim_{k 鈫� 鈭�} \frac{k^{p}}{k \ln k} = \lim_{k 鈫� 鈭�} \frac{k^{p - 1}}{\ln k} = \lim_{k 鈫� 鈭�} \frac{(p - 1) k^{p - 2}}{\frac{1}{k}} (using' s l hospital rule) = \lim_{k 鈫� 鈭�} \frac{(p - 1)}{k^{1 - p}} = 0$

Step 3. Limit comparison test

$The limit comparison test states that for and be two series with positive terms then If \lim_{k 鈫� 鈭�} \frac{a_{k}}{b_{k}} = L Where L is any positive real number then either both converges or diverges . If \lim_{k 鈫� 鈭�} \frac{a_{k}}{b_{k}} = 0 and {鈭�}_{k = 1}^{鈭�} b_{k} converges then {鈭�}_{k = 1}^{鈭�} a_{k} converges . If \lim_{k 鈫� 鈭�} \frac{a_{k}}{b_{k}} = 鈭� and {鈭�}_{k = 1}^{鈭�} b_{k} diverges then {鈭�}_{k = 1}^{鈭�} a_{k} diverges .$

Step 4. Result

$The value of the expression is \lim_{k 鈫� 鈭�} \frac{1 / k lnk}{1 / k^{p}} = 0 The p - series is divergent for 0 < p < 1 . The limit comparison test if the ratio is zero states that if \lim_{k 鈫� 鈭�} \frac{a_{k}}{b_{k}} and {鈭�}_{k = 1}^{鈭�} b_{k} converges, then {鈭�}_{k = 1}^{鈭�} a_{k} converges The limit comparison test fails to give any information about the divergence of the series {鈭�}_{k = 2}^{鈭�} \frac{1}{k lnk}$

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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 ${鈭�}_{k = 2}^{鈭�} \frac{1}{k \log k}$

Short Answer

Step by step solution

Step 1. Given

Step 2.Finding the value of the expression

Step 3. Limit comparison test

Step 4. Result

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91影视

Let 0 < p < 1. Evaluate the limitlimk鈫�鈭�1/klnk1/kpExplain 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�1klogk

Short Answer

Step by step solution

Step 1. Given

Step 2.Finding the value of the expression

Step 3. Limit comparison test

Step 4. Result

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Geometry

Calculus

Discrete Mathematics

Applied Mathematics

Decision Maths

Study anywhere. Anytime. Across all devices.

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 ${鈭�}_{k = 2}^{鈭�} \frac{1}{k \log k}$