How to Use Comparison Test for Convergence?

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Homework Help Overview

The discussion revolves around determining the convergence or divergence of the series \(\sum_{n=0}^{\infty}{\frac{1}{\sqrt{n+1}}}\) using the comparison test and potentially the limit comparison test.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the comparison test and the limit comparison test, questioning how to appropriately select a comparison series. There is mention of comparing the series to \(\sum_{n=0}^{\infty}{\frac{1}{\sqrt{n}}}\) and concerns about its divergence.

Discussion Status

Participants are exploring various methods for applying the comparison tests, with some suggesting alternatives and others providing guidance on the use of the limit comparison test. There is an acknowledgment of potential issues with specific terms in the series, but productive suggestions are being made.

Contextual Notes

Some participants express confusion regarding the behavior of the series at \(n=0\) and the implications for convergence, while others emphasize that the first term does not affect the overall convergence of the series.

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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