Question

Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \sqrt{n}}{\sqrt[3]{n}} $$

Short answer

The series converges.

Step-by-step solution

Simplify The Series

Firstly, simplify the series. The n-th term of the series is \( \frac{(-1)^{n+1} \sqrt{n}}{\sqrt[3]{n}} \). This simplifies to \( (-1)^{n+1} {n}^{1/6} \).

Apply The Alternating Series Test

To apply the Alternating Series Test, we need to show that the absolute value of the series terms is decreasing and that the limit of the series terms as n approaches infinity is zero. Firstly, to show that the terms are decreasing, take two subsequent terms \(a_n\) and \(a_{n+1}\). Their ratio \(a_{n+1}/a_n = (n+1)^{1/6}/n^{1/6}\) is increasing as n increases, which means that the terms of the original series are decreasing. Secondly, take the limit of the n-th term as n approaches infinity; we find that the limit is 0.

Conclusion on Convergence

Since the two conditions of the Alternating Series Test are satisfied, our series is convergent.