Answered: (-1)" /n n +1 n=1

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.3: Geometric Sequences

Problem 49E

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Question

Determine if the seriesconverges absolutely, conditionally, or diverges.

Expert Solution

Step 1

Given series is $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} \sqrt{n}}{n + 2}$

Let $a_{n} = \frac{\sqrt{n}}{n + 2}$

Since, the sequence $\{\frac{\sqrt{n}}{n + 2}\}$ is monotonically decreasing and $\lim_{n \to \infty} \frac{\sqrt{n}}{n + 2} = 0$

So, by Leibnitz test $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} \sqrt{n}}{n + 2}$ is convergent .

Step 2

Now, consider

$a_{n} = \frac{\sqrt{n}}{n + 2} = \frac{1}{\sqrt{n} (1 + \frac{2}{n})}$

take $b_{n} = \frac{1}{\sqrt{n}}$

Then,

$\lim_{n \to \infty} \frac{a_{n}}{b_{n}} = \lim_{n \to \infty} \frac{1}{1 + \frac{2}{n}} = 1$ (finite)

So, by comparison test $\sum_{n = 1}^{\infty} a_{n}, \sum_{n = 1}^{\infty} b_{n}$ converges or diverges together .

Since the series $\sum_{n = 1}^{\infty} b_{n} = \sum_{n = 1}^{\infty} \frac{1}{\sqrt{n}}$ diverges by P test for series . So, the series $\sum_{n = 1}^{\infty} a_{n}$ also diverges .

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