Let 0 < p < 1. Evaluate the limit $\lim_{k \to \infty} \frac{1 / k \ln k}{1 / k^{p}}$
Explain why we cannot use a p-series with 0 < p < 1 in a limit comparison test to verify the divergence of the series $\sum_{k = 2}^{\infty} \frac{1}{k \log k}$

Question

Let 0 < p < 1. Evaluate the limit $\lim_{k \to \infty} \frac{1 / k \ln k}{1 / k^{p}}$
Explain why we cannot use a p-series with 0 < p < 1 in a limit comparison test to verify the divergence of the series $\sum_{k = 2}^{\infty} \frac{1}{k \log k}$

Accepted Answer

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

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Calculus

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Short Answer

Let 0 < p < 1. Evaluate the limit $\lim_{k \to \infty} \frac{1 / k \ln k}{1 / k^{p}}$
Explain why we cannot use a p-series with 0 < p < 1 in a limit comparison test to verify the divergence of the series $\sum_{k = 2}^{\infty} \frac{1}{k \log k}$

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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Calculus

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Short Answer

Let 0 < p < 1. Evaluate the limit limk→∞1/k lnk1/kp Explain why we cannot use a p-series with 0 < p < 1 in a limit comparison test to verify the divergence of the series ∑k=2∞1k log k

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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Let 0 < p < 1. Evaluate the limit $\lim_{k \to \infty} \frac{1 / k \ln k}{1 / k^{p}}$
Explain why we cannot use a p-series with 0 < p < 1 in a limit comparison test to verify the divergence of the series $\sum_{k = 2}^{\infty} \frac{1}{k \log k}$