# How to apply comparison test?

1. Apr 15, 2013

### Turion

1. The problem statement, all variables and given/known data

Is the series convergent or divergent? $$\sum_{n=0}^{\infty}{\frac{1}{\sqrt{n+1}}}$$

2. Relevant equations

I can use any test but wolfram alpha says that it is divergent by comparison test.

3. The attempt at a solution

How do I apply comparison test?

I can compare it to: $$\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ \sqrt { n } } }$$ but the second series is greater than the series in the question and the second series is divegent. :(

2. Apr 15, 2013

### LCKurtz

Have you had the "generalized comparison test"? That's what you want.

3. Apr 15, 2013

### Turion

You mean the limit comparison test? Yes, I've learned both the comparison test and the limit comparison test.

Currently trying to use the second series for limit comparison test.

4. Apr 15, 2013

### LCKurtz

Yes. That should work.

5. Apr 15, 2013

### Turion

Wait. I found an easier method. Just let k=n+1.

6. Apr 15, 2013

### LCKurtz

Yes, that's pretty slick. Still, the limit comparison test is your friend and you want to be adept at recognizing when and how to use it. Your idea won't work on lots of similar problems that are just a little bit more complicated.

7. Apr 15, 2013

### Turion

I just realized that my second limit does not make sense since when n=0, you get 1/0.

What do I use for limit comparison test? :(

8. Apr 15, 2013

### LCKurtz

That doesn't matter. It only matters what happens at the tail end of the series. Just leave off the first term; it won't affect convergence.

Use your original $\frac 1{\sqrt n}$ for the limit comparison test.

9. Apr 15, 2013

### Turion

Haha. Got it. Thank you!