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Chapter 8: Problem 14

Use the Limit Comparison Test to determine whether the series is convergent or divergent. \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}+2}\)

Short Answer

Expert verified
The series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}+2}\) can be compared to the series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\) using the Limit Comparison Test. Calculate the limit of the ratio of the two series as n approaches infinity: \(\lim_{n\to\infty} \frac{\frac{1}{\sqrt{n}+2}}{\frac{1}{\sqrt{n}}} = \lim_{n\to\infty} \frac{\sqrt{n}}{\sqrt{n}+2} = 1\). Since the limit is 1, both series have the same behavior. The comparison series is a divergent p-series with \(p = \frac{1}{2}\), thus the given series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}+2}\) also diverges.

Step by step solution

01

Choose a comparison series

Choose a series that has a similar form to the given series and would make it easy to compute the limit. A common first choice for comparison series when dealing with a radical is to look at the power term within the radical. So the series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\) would be a good comparison series.
02

Calculate the limit of the ratio

Find the limit of the ratio between the terms of the given series and the comparison series: \(\lim_{n\to\infty} \frac{\frac{1}{\sqrt{n}+2}}{\frac{1}{\sqrt{n}}}\)
03

Simplify the ratio

Simplify the limit expression by combining the fractions: \(\lim_{n\to\infty} \frac{1}{\sqrt{n}+2} \cdot \frac{\sqrt{n}}{1} = \lim_{n\to\infty} \frac{\sqrt{n}}{\sqrt{n}+2}\)
04

Compute the limit

Calculate the limit as n approaches infinity in the simplified expression: \(\lim_{n\to\infty} \frac{\sqrt{n}}{\sqrt{n}+2} = 1\)
05

Analyze the result

Since the limit is 1 (finite and greater than 0), the given series and the comparison series either both converge or both diverge. The comparison series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\) is a p-series with \(p = \frac{1}{2}\). Since \(p\) is less than or equal to 1, the comparison series diverges by the p-series test.
06

Draw the conclusion

Because the comparison series diverges and the limit result shows that their behavior is the same, the series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}+2}\) also diverges based on the Limit Comparison Test.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convergent Series
A convergent series is a sequence of numbers where the sum of its terms approaches a specific, finite number as you add more and more terms. This happens when the terms of the series get smaller and smaller and eventually become negligible.
Some important characteristics include:
  • The sum of the series approaches a finite limit.
  • Often involves a term that decreases quickly enough to reach that limit.
Think of it like a long road trip where the destination gets closer with every mile, and you stop when you reach the endpoint.
Using tests like the Limit Comparison Test helps determine if a series converges by comparing it to a known convergent series.
Divergent Series
Unlike convergent series, a divergent series keeps growing larger and never approaches a finite limit. Each added term makes the sum larger or causes it to fluctuate without settling.
Characteristics of divergent series include:
  • The sum does not converge to a specific number.
  • Terms might decrease but not quickly enough to reach a finite limit.
Imagine continuously walking on a circular path without ever arriving at a destination.
The Limit Comparison Test helps identify divergence by comparing with a known divergent series, providing insight into the behavior of a series like \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}+2}\).
P-Series Test
The p-series test is a method used to determine convergence or divergence by examining series in the form \(\sum_{n=1}^{\infty} \frac{1}{n^p}\). The series converges if the exponent \(p\) is greater than 1 and diverges if \(p\) is less than or equal to 1.
Key features of the p-series test:
  • Converges for \(p > 1\). For example, \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) converges.
  • Diverges for \(p \leq 1\). For example, \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\), where \(p = \frac{1}{2}\), diverges.
This test provides a simple way to analyze series and understand their long-term behavior, simplifying complex evaluations like in the original exercise, where the divergent nature of the comparison series was confirmed using the p-series test.

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Most popular questions from this chapter

Use the power series representations of functions established in this section to find the Taylor series of \(f\) at the given value of \(c .\) Then find the radius of convergence of the series. \(f(x)=\frac{x^{2}}{x^{2}-1}, \quad c=0\)

Show that \((1+x)^{n}>1+n x\) for all \(x>0\) and \(n>1\).

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. Suppose that \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) for \(x\) in \((-R, R)\), where \(R>0\) and \(f\) is odd. Then \(a_{2 n}=0\) for \(n \geq 0\).

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