Introduction to Analyzing Series Convergence with Variable Alpha

In this article, we will explore the convergence behavior of the series defined by:

[sum_{n1}^infty frac{(-1)^n}{n^{alpha}}]

We will discuss the cases where the series converges absolutely, conditionally, or diverges, and then extend this analysis to a related series:

[sum_{n1}^infty frac{n-1^n}{1n^{2p}}]

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

For the given series, we need to analyze its convergence properties under different conditions. This involves two parts:

When the series converges absolutely, converges conditionally, or diverges. When comparing the given series with another series to determine its convergence behavior.

Part 1: Convergence Analysis

Firstly, let's consider the series (sum_{n1}^infty frac{(-1)^n}{n^{alpha}}).

The series converges absolutely when the following limit is zero:

[lim_{n to infty} frac{1}{n^{alpha}} 0]

This is true if and only if (alpha > 0).

For the series to converge conditionally, it must converge by the Alternating Series Test (ASET), which requires two conditions:

The absolute value of the terms must be eventually decreasing. The limit of the terms must approach zero.

The limit condition is automatically satisfied if (alpha > 0). Now, let's check if the terms are eventually decreasing:

[frac{d}{dn} left(frac{n}{1 n^{2alpha}}right) frac{1 - 2alpha n^{2alpha}}{(1 n^{2alpha})^2}]

The derivative equals zero when (n sqrt{2alpha}). For (n > sqrt{2alpha}), the derivative is negative, ensuring that the terms are eventually decreasing.

Hence, the series (sum_{n1}^infty frac{(-1)^n}{n^{alpha}}) converges when (alpha > 0).

Part 2: Comparison with Another Series

Next, we analyze the series (sum_{n1}^infty frac{n-1^n}{1n^{2p}}). We will compare this with the related series:

[sum_{n1}^infty frac{n}{1 n^{2p}}]

By the Limit Comparison Test, the convergence of (sum_{n1}^infty frac{n}{1 n^{2p}}) is equivalent to the convergence of (sum_{n1}^infty frac{1}{n^{2p-1}}), which is a p-series.

The p-series (sum_{n1}^infty frac{1}{n^{2p-1}}) converges if and only if (2p-1 > 1), which simplifies to (p > 1).

Therefore, the original series (sum_{n1}^infty frac{n-1^n}{1n^{2p}}) is absolutely convergent when (p > 1) and conditionally convergent when (frac{1}{2} leq p leq 1).

Conclusion

In summary, the series (sum_{n1}^infty frac{(-1)^n}{n^{alpha}}) converges absolutely when (alpha > 0) and conditionally converges when (0 1) and conditionally convergent when (frac{1}{2} leq p leq 1).