Chapter 7: Q. 4 (page 631)

Use the comparison test to explain why the series $\sum_{k = 1}^{\infty} \frac{1}{\sqrt[α]{k}}$ diverges when $α$ is an integer greater than $1$

Short Answer

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The series $\sum_{k = 1}^{\infty} \frac{1}{\sqrt[α]{k}}$ diverges when $α$ is greater than $1$

Step by step solution

Step 1. Given information

An series is given as $\sum_{k - 1}^{\infty} \frac{1}{\sqrt[α]{k}}$

Step 2. Applying comparison test

Terms of the given series are positive.

Now the series $\sum_{k = 1}^{\infty} b_{k}$ can be written as

$\sum_{k = 1}^{\infty} b_{k} = \frac{1}{\sqrt[a]{k}}$

After that the ratio is $\lim_{k \to \infty} \frac{a_{k}}{b_{k}}$ can be written as:

$\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = \lim_{k \to \infty} \frac{\frac{1}{\sqrt[a]{k}}}{\frac{1}{\sqrt[a]{k}}} = \lim_{i \to \infty} 1 = 1$

The value of $\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = 1$ is a non-zero finite number.

Considering the conditions $a > 1 & \frac{1}{a} < 1$ , the series $\sum_{k = 1}^{\infty} b_{k} = \sum_{k = 1}^{\infty} \frac{1}{k^{1 / a}}$ is divergent by p-series. Therefore the series $\sum_{k = 1}^{\infty} a_{k}$ is also divergent

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Use the comparison test to explain why the series $\sum_{k = 1}^{\infty} \frac{1}{\sqrt[α]{k}}$ diverges when $α$ is an integer greater than $1$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Applying comparison test

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Use the comparison test to explain why the series ∑k=1∞1kαdiverges when αis an integer greater than 1

Short Answer

Step by step solution

Step 1. Given information

Step 2. Applying comparison test

One App. One Place for Learning.

Most popular questions from this chapter

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Use the comparison test to explain why the series $\sum_{k = 1}^{\infty} \frac{1}{\sqrt[α]{k}}$ diverges when $α$ is an integer greater than $1$