# Homework Help: (conditional) convergence

1. Nov 26, 2014

### nuuskur

1. The problem statement, all variables and given/known data
Consider the series: $$\sum\limits_{k=17}^\infty (-1)^{k}(\sqrt{k-3}-\sqrt{k-5})$$

2. Relevant equations

3. 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?

2. Nov 26, 2014

### nuuskur

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

3. Nov 26, 2014

### Ray Vickson

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 $k \to \infty$.