# Series root or ratio test?

## 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..?

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

## The Attempt at a Solution

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Mark44
Mentor
$$\sqrt{k} - \sqrt{k - 1} = \frac{(\sqrt{k} - \sqrt{k - 1})(\sqrt{k} + \sqrt{k - 1})}{\sqrt{k} + \sqrt{k - 1}}$$

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?

Mark44
Mentor
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

Mark44
Mentor
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?

Mark44
Mentor
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

Mark44
Mentor
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.