How to Use Comparison Test for Convergence?

Turion
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Homework Statement



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

Homework Equations



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

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. :(
 
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Turion said:

Homework Statement



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

Homework Equations



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

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. :(

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

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.
 
Yes. That should work.
 
Wait. I found an easier method. Just let k=n+1.
 
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.
 
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? :(
 
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.
 
Haha. Got it. Thank you!
 

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