Convergence Test for the Series 1/√n sin(1/n): An In-Depth Analysis

In this article, we will explore the detailed process of determining the convergence of the series 1/√n sin(1/n). We will utilize both the limit comparison test and the comparison test to ascertain its behavior. Understanding these methods is crucial for advanced calculus and real analysis, providing a deeper insight into the convergence properties of series.

Introduction

The series in question is 1/√n sin(1/n). Determining whether such a series converges or diverges involves rigorous mathematical analysis. The two primary tests we will employ are the limit comparison test and the comparison test. Both methods are widely used in mathematical analysis and provide valuable insights into the behavior of series.

Limit Comparison Test

To apply the limit comparison test, we first choose a suitable series to compare with. Here, we select the series b_n 1/√n. This choice is based on the fact that the behavior of sin(1/n) as n approaches infinity is similar to that of 1/n.

The test requires us to compute the following limit:

limn→∞ (a_n / b_n) limn→∞ (sin(1/n) / (1/n))

Using L'Hopital's rule, we evaluate this limit:

limn→∞ (sin(1/n) / (1/n)) limn→∞ ((-cos(1/n) * (-1/n^2)) / (-1/n^2)) limn→∞ cos(1/n) 1

Since the limit is a positive finite number (1), the limit comparison test tells us that the series ∑ a_n and ∑ b_n either both converge or both diverge. To confirm, we must check if the series ∑ b_n converges.

Given that ∑ b_n is a p-series with p 1/2 and p > 1, it converges. Therefore, by the limit comparison test, the series ∑ a_n also converges.

Comparison Test

The comparison test states that if 0 ≤ a_n ≤ b_n for all n, and if ∑ b_n converges, then ∑ a_n also converges.

For the series 1/√n sin(1/n), we can establish the following inequality:

1/√n sin(1/n) ≤ 1/√n

This inequality holds because -1 ≤ sin(1/n) ≤ 1. Therefore, multiplying both sides by 1/√n preserves the inequality:

1/√n * -1 ≤ 1/√n * sin(1/n) ≤ 1/√n * 1

This simplifies to:

-1/√n ≤ 1/√n sin(1/n) ≤ 1/√n

Since ∑ 1/√n is a p-series with p 1/2 and p > 1, it converges. By the comparison test, if 1/√n sin(1/n) is bounded above by a convergent series, then 1/√n sin(1/n) also converges.

Conclusion

Both the limit comparison test and the comparison test confirm that the series 1/√n sin(1/n) converges. This analysis demonstrates the power of these convergence tests in determining the behavior of complex series involving trigonometric functions.

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