Problem 14 - Single Variable Calculus

Question

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

Step-by-Step Solution

Answer

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 1: 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.

Step 2: Calculate the limit of the ratio

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

Step 3: Simplify the ratio

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

Step 4: Compute the limit

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

Step 5: 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.

Step 6: 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.

Key Concepts

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.