Determining Convergence of the Series Using the Alternating Series Test and Limit Comparison Test

In this article, we will explore how to determine the convergence of the series (sum_{n1}^{infty} frac{-1^n}{n^{alpha}}). This involves understanding the conditions under which the series converges absolutely, conditionally, or diverges. We will also discuss how to apply the Alternating Series Test and the Limit Comparison Test to solve related problems.

Convergence of the Series (sum_{n1}^{infty} frac{-1^n}{n^{alpha}})

Let's start with the general series (sum_{n1}^{infty} frac{-1^n}{n^{alpha}}). For the series to converge, it must satisfy certain conditions based on the value of (alpha).

Case 1: Absolute Convergence

Firstly, we need to determine when the series converges absolutely. This means that the series (sum_{n1}^{infty} left| frac{-1^n}{n^{alpha}} right| sum_{n1}^{infty} frac{1}{n^{alpha}}) converges. This series is a p-series, and it converges if and only if (alpha > 1). Hence, for absolute convergence of the original series, (alpha > 1).

Case 2: Conditional Convergence

Next, we consider the case where the series converges conditionally. For this, we need to check when the series converges but not absolutely. This involves using the Alternating Series Test.

Alternating Series Test

The Alternating Series Test states that the series (sum_{n1}^{infty} a_n) with (a_n frac{1}{n^{alpha}}) converges if:

The sequence (a_n) is monotonically decreasing. (lim_{nto infty} a_n 0).

We have already discussed that for the series to converge absolutely, we need (alpha > 1). However, we want to find out when the series converges conditionally. We need to ensure that:

The sequence (a_n frac{1}{n^{alpha}}) is monotonically decreasing, which is true if (alpha > 0). (lim_{nto infty} frac{1}{n^{alpha}} 0), which is true for any (alpha > 0).

Now, we need to check when the series is conditionally convergent. This happens when (1/2 alpha 1). Therefore, the series is conditionally convergent for (1/2 alpha leq 1).

Special Case: Limit Comparison Test

We also apply the Limit Comparison Test to the related series (sum_{n1}^{infty} frac{n}{1 n^{2p}}). For large (n), we can approximate:

(frac{n}{1 n^{2p}} sim frac{n}{n^{2p}} frac{1}{n^{2p-1}}).

The Limit Comparison Test states that if the limit (lim_{nto infty} frac{frac{n}{1 n^{2p}}}{frac{1}{n^{2p-1}}} L) exists and is not zero, then (sum_{n1}^{infty} frac{n}{1 n^{2p}}) and (sum_{n1}^{infty} frac{1}{n^{2p-1}}) either both converge or both diverge. Since the p-series converges if and only if (2p - 1 1), we have (p 1) for absolute convergence and (1/2 p leq 1) for conditional convergence.

Conclusion

Summarizing our findings:

The series (sum_{n1}^{infty} frac{-1^n}{n^{alpha}}) converges if (alpha 1/2). The series converges absolutely if (alpha 1). The series converges conditionally if (1/2 alpha leq 1). The series diverges if (alpha 1/2).

Thus, the value of (alpha) determines the nature of the convergence of the series. Understanding these concepts is crucial for anyone dealing with series in mathematical analysis.