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).