# Series root or ratio test?

• zeion

## Homework Statement

Determine whether the series converges or diverges.

$$\sum (\sqrt {k} - \sqrt {k - 1})^k$$

## The Attempt at a Solution

$$(a_k)^\frac{1}{k} = \sqrt{k} - \sqrt{k - 1}$$

What do I do here..?

$$= \frac{1}{\sqrt{k} + \sqrt{k-1}} \to 0 ?$$

$$\sqrt{k} - \sqrt{k - 1} = \frac{(\sqrt{k} - \sqrt{k - 1})(\sqrt{k} + \sqrt{k - 1})}{\sqrt{k} + \sqrt{k - 1}}$$

So that is
$$\sqrt{k} - \sqrt{k - 1} = \frac{(\sqrt{k} - \sqrt{k - 1})(\sqrt{k} + \sqrt{k - 1})}{\sqrt{k} + \sqrt{k - 1}} = \frac {\sqrt{k}^2 - \sqrt{k-1}^2}{\sqrt{k} + \sqrt{k - 1}} = \frac{1}{\sqrt{k} + \sqrt{k-1}}$$

Then

$$\frac{1}{\sqrt{k} + \sqrt{k-1}} < \frac{1}{\sqrt{k}} = ( \frac{1}{k})^{\frac{1}{2}$$? then converges?

You are apparently confusing yourself. You started with the root test, not the comparison test.

BTW, the series whose general term is 1/sqrt(k) diverges, but that's not relevant to what you're doing.

Ok so I go here $$\frac{1}{\sqrt{k} + \sqrt{k-1}}$$ and then I'm kinda stuck

Take the limit as k --> infinity. What do you get? Why is this limit important to you? What did you start out doing in your first post?

I get 0? So the ratio is < 1 so its converges?

The limit is < 1. Try not to confuse yourself into think you are working with the ratio test - here it's the root test, so the fact that you are finding the limit of a fraction is not relevant.

Ok so since $$a_k < \mu^k$$ and $$\mu^k$$ converges $$a_k$$converges

I don't know -- what's $\mu$? That's the first time it has appeared in this thread.

zeion, you need to step back and take a bigger-picture view of what you're doing. You seem to be getting lost in minute details, and losing track of the purpose of the details.

1. What test are you using?
2. When you use this test, what result indicates that the series you're testing converges?
What result indicates that the series diverges?
3. What result did you get?
4. What can you conclude about the series you are testing?