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Chapter 36: Chapter 36

Problem 1135

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Establish the convergence or divergence of the series: $[1 /(1+\sqrt{1})]+[1 /(1+\sqrt{2})]+[1 /(1+\sqrt{3})]+[1 /(1+\sqrt{4})]+\ldots$

Short Answer

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The series \(\sum_{n=1}^{\infty} \frac{1}{1 + \sqrt{n}}\) diverges, as demonstrated by the Comparison Test with the harmonic series \(\sum_{n=1}^{\infty} \frac{1}{n}\), which also diverges. The inequality \(\frac{1}{1 + \sqrt{n}} \geq \frac{1}{n}\) holds for all \(n \geq 1\), showing that the terms of the given series are larger than those of the harmonic series.
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TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Write the series and the harmonic series for comparison

We have the given series: \[ \sum_{n=1}^{\infty} \frac{1}{1 + \sqrt{n}}\] And the harmonic series: \[ \sum_{n=1}^{\infty} \frac{1}{n} \]

Step 2: Compare the terms of the series

In order to compare the terms of the given series and the harmonic series, we will analyze the inequality: \[ \frac{1}{1 + \sqrt{n}} \geq \frac{1}{n} \] For all \(n \geq 1\), it is easy to see that \(\sqrt{n} \geq 1\), which implies that \(1 + \sqrt{n} \geq 2\). Therefore, \( n \geq 1 + \sqrt{n} \). Consequently, we have: \[ \frac{1}{n} \leq \frac{1}{1 + \sqrt{n}}\]

Step 3: Apply the Comparison Test

Since the terms of our series are larger than the terms of the harmonic series, which is known to diverge, then by the Comparison Test, our series also diverges. Thus, the series: \[ \sum_{n=1}^{\infty} \frac{1}{1 + \sqrt{n}}\] diverges.

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Most popular questions from this chapter

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