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Expert-verified Found in: Page 625 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Explain why the integral test may be used to analyze the given series and then use the test to determine whether the series converges or diverges. $\sum _{k=2}^{\mathrm{\infty }} \frac{1}{k\sqrt{\mathrm{ln}k}}$

Ans: The series $\sum _{k=2}^{\mathrm{\infty }} \frac{1}{k\sqrt{\mathrm{ln}k}}$ is divergent.

See the step by step solution

## Step 1. Given information.

given,

$\sum _{k=2}^{\mathrm{\infty }} \frac{1}{k\sqrt{\mathrm{ln}k}}$

## Step 2. The objective is to explain why the integral test is used to determine the convergence or divergence of the series and use the test to determine the convergence or divergence of the series.

Consider function $f\left(x\right)=\frac{1}{x\sqrt{\mathrm{ln}x}}$.

The function $f\left(x\right)=\frac{1}{x\sqrt{\mathrm{ln}x}}$ is continuous, decreasing, with positive terms. Therefore, all the conditions of the integral test are fulfilled. So, the integral test is applicable.

## Step 3. Consider the integral ∫x=2∞ f(x)dx=∫x=2∞ 1xln⁡xdx

Therefore,

$\begin{array}{r}{\int }_{x=2}^{\mathrm{\infty }} f\left(x\right)dx=\underset{k\to \mathrm{\infty }}{lim} {\int }_{x=2}^{k} \frac{1}{x\sqrt{\mathrm{ln}x}}dx\\ =\underset{k\to \mathrm{\infty }}{lim} {\int }_{u=\mathrm{ln}2}^{\mathrm{ln}k} \frac{1}{\sqrt{u}}du\text{(Put}\mathrm{ln}x=u,⇒\frac{1}{x}dx=du\text{)}\\ =\underset{k\to \mathrm{\infty }}{lim} \left[2\sqrt{u}{\right]}_{\mathrm{ln}2}^{\mathrm{ln}k}\\ =2\underset{k\to \mathrm{\infty }}{lim} \left[\sqrt{\mathrm{ln}k}-\mathrm{ln}2\right]\text{(Substitution)}\\ =2\left(\mathrm{\infty }-\mathrm{ln}2\right)\text{(Take limit)}\\ =\mathrm{\infty }\end{array}$

## Step 4. Thus, the value of the integral is ∫x=2∞ 1xln⁡xdx=∞

The integral converges. Therefore, the series $\sum _{k=2}^{\mathrm{\infty }} \frac{1}{k\sqrt{\mathrm{ln}k}}$ is divergent.

Hence, by integral test, the series $\sum _{k=2}^{\mathrm{\infty }} \frac{1}{k\sqrt{\mathrm{ln}k}}$ is divergent

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