Converge or diverge

Recall the alternating series test for convergence :

If $a_n$ decreases monotonically and $\displaystyle \lim_{n \ \to \ \infty} a_n = 0$ , then $\displaystyle \sum_{n \ = \ k}^{\infty} \left(- 1\right)^na_n$ converges.

$\displaystyle \begin{aligned} \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}} + \cdots = \sum_{n \ = \ 1}^{\infty} \frac{(- 1)^{n + 1}}{\sqrt{n}} \end{aligned}$

Let $\displaystyle a_n = \frac{1}{\sqrt{n}}$ .

Condition 1: $a_n > a_{n + 1}$

$\displaystyle \begin{aligned} n & < n + 1 \\ \implies \sqrt{n} & < \sqrt{n + 1} & \small \color{#3D99F6} \forall \ n \geq 0 \\ \implies \frac{1}{\sqrt{n}} & > \frac{1}{\sqrt{n + 1}} & \small \color{#3D99F6} \forall \ n \geq 1 \end{aligned}$

Condition 2: $\displaystyle \lim_{n \ \to \ \infty} a_n = 0$

$\displaystyle \lim_{n \ \to \ \infty} \frac{1}{\sqrt{n}} = \lim_{n \ \to \ \infty} \frac{1}{\infty} = 0$

Since both conditions are satisfied, $\displaystyle \sum_{n \ = \ 1}^{\infty} \frac{(- 1)^{n + 1}}{\sqrt{n}}$ converges by the alternating series test.