Testing conditional convergence of alternating series with ratio test limit 1

[nuuskur]

Homework Statement

Consider the series: $$\sum\limits_{k=17}^\infty (-1)^{k}(\sqrt{k-3}-\sqrt{k-5})$$

Homework Equations

The Attempt at a Solution

First, I will attempt to determine whether it is absolutely convergent:
$$\lim\limits_{k\to\infty} \left(\sqrt{k-3}-\sqrt{k-5}\right) = 0$$
Since the limit is 0 I can continue by applying the d'Alembert's ratio test:
$$\lim\limits_{k\to\infty} \frac{\sqrt{k-2}-\sqrt{k-4}}{\sqrt{k-3}-\sqrt{k-5}} = 1$$
If the ratio is 1, we don't know whether it converges or diverges. If the limit of the ratio test is 1, how do we proceed?
According to Leibniz's principle: if $a_k \geq a_{k+1}, \forall k\in [17,\infty), \lim\limits_{k} a_k = 0$ , then the series $\sum (-1)^{k}a_k$ converges.
Is $a_k \geq a_{k+1}$?Let's try k=17: $\sqrt{14} - \sqrt{12} \geq \sqrt{15} - \sqrt{13}$? Yes, it is. Would I have to show via induction that this is always true? (should be quite obvious, why, though).
What happens with the ratio test, then? It didn't pass, can we conclude the series is absolutely convergent?

 

[nuuskur]

Okay, this series is not absolutely convergent, but it is conditionally convergent (the summands' values are decreasing).

 

[Ray Vickson]

Okay, this series is not absolutely convergent, but it is conditionally convergent (the summands' values are decreasing).

This is not quite enough; you also need to show that the magnitude of the kth term $a_k = \sqrt{k-3}-\sqrt{k-5}$ goes to 0 in the limit $k \to \infty$.