\( \sum _ { n = 2 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { [ \sqrt[ 3 ] { n } + ( - 1 ) ^ { n } ] ^ { \alpha } } \)

To understand the convergence of the series \( \sum_{n=2}^{\infty} \frac{(-1)^n}{[\sqrt[3]{n} + (-1)^n]^{\alpha}} \), we can analyze it using the Alternating Series Test. For convergence, we need to check that the terms \( b_n = \frac{1}{[\sqrt[3]{n} + (-1)^n]^{\alpha}} \) decrease monotonically and tend to zero as \( n \) approaches infinity. As \( n \) becomes very large, \( \sqrt[3]{n} \) dominates the \( (-1)^n \) term, and we find: 1. For even \( n \): \( b_{2k} = \frac{1}{[\sqrt[3]{2k} + 1]^\alpha} \) 2. For odd \( n \): \( b_{2k+1} = \frac{1}{[\sqrt[3]{2k+1} - 1]^\alpha} \) Both \( b_{2k} \) and \( b_{2k+1} \) tend to \( 0 \) as \( n \) goes to infinity, thus satisfying the necessary condition for convergence. However, for sufficient convergence, we also need the limit of \( b_n \) as \( n \to \infty \) to decrease. This involves examining the decay rate dictated by \( \alpha \). In general, this series converges if \( \alpha > 1 \), mimicking convergence behavior similar to that of a p-series. So, make sure to check the value of \( \alpha \) for your specific needs!