Question

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

Short answer

The series \(\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\) is convergent, as established by the Alternating Series Test.

Step-by-step solution

Identify the series as an alternating series

The series can be written as \(\sum_{n=1}^{\infty} (-1)^{n}b_n\), where \(b_n\) is positive and \(b_n = \frac{1}{\sqrt{n}}\). The factor of \((-1)^n\) makes the series alternate between negative and positive terms.

Check if \(b_{n+1} \leq b_n\) for all \(n\)

We need to show that \(\frac{1}{\sqrt{n+1}} \leq \frac{1}{\sqrt{n}}\). It is equivalent to \(\sqrt{n} \leq \sqrt{n+1}\) which is true for all positive \(n\). Hence, the series' terms \(b_n\) are decreasing.

Check if \(\lim_{n \to \infty} b_n = 0\)

The limit of \(b_n\) as \(n\) approaches infinity is \(\lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0\). This comes from the fact that as \(n\) becomes larger, \(\sqrt{n}\) also becomes larger, so the whole fraction approaches 0.

Apply the Alternating Series Test

Since the two conditions of the Alternating Series Test are met, we can conclude that the series \(\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\) is convergent.