Is the series covergent or divergent?

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    Divergent Series
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Discussion Overview

The discussion revolves around determining the convergence or divergence of the series \(\sum \frac{2}{\sqrt{n}+1}\) and a related series \(\sum \frac{1}{\sqrt{n}+\sqrt{n+1}}\). Participants explore various tests for convergence and share their reasoning and challenges in applying these tests.

Discussion Character

  • Debate/contested
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant questions the convergence of the series \(\sum \frac{2}{\sqrt{n}+1}\) and mentions that the divergence test yields a result of 0, leading to confusion about which test to apply.
  • Another participant suggests comparing the series to \(\sum \frac{1}{n}\) as a potential method for analysis.
  • A participant expresses uncertainty about the correct answer being convergent, prompting a clarification on the limits of the series.
  • A later reply indicates that the limit comparison test with \(\sum \frac{1}{n}\) was successful in determining divergence.
  • One participant acknowledges the divergence of the harmonic series \(\sum \frac{1}{n}\) and corrects their earlier assertion about the series being convergent.
  • The discussion shifts to a new series \(\sum \frac{1}{\sqrt{n}+\sqrt{n+1}}\), with a participant expressing difficulty in finding a suitable comparison series.
  • Another participant suggests considering the behavior of the terms as \(n\) becomes large to identify an appropriate comparison.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the convergence of the initial series, with some asserting divergence and others initially suggesting convergence. The discussion on the second series remains unresolved, with participants exploring potential comparison series without agreement on a conclusion.

Contextual Notes

Participants express uncertainty regarding the application of various convergence tests and the choice of comparison series, indicating that assumptions about limits and series behavior are not fully established.

shaiqbashir
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is the series covergent or divergent?

I want to know that is the following series convergent or divergent??

\sum \frac{2}{\sqrt{n}+1}


when i apply divergent test to it, it comes equal to 0 , it means that divergent test gets failed. then how to solve it?

which test i should apply?

the correct answer in my copy is written as COnvergent, but I am not getting a convergent answer.

please help me as soon as possible!

Thanks in advance!
 
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Compare it to \sum 1/n.
 
shaiqbashir said:
the correct answer in my copy is written as COnvergent, but I am not getting a convergent answer.
Are you sure about that? Perhaps you should specify the limits, or may we assume from 0 (or 1?) to infinity?
 
Thanks for ur Help StatusX!

I applied Limit comparison test and it works with series 1/n as Bn.

thanks a lot once again!
 
But you know that the harmonic series (1/n) diverges, right?
 
okz okz! I am sorry for some mistakes.

the correct answer of the above series is this that the series is DIVERGENT not convergent. it was my mistake!
So that means that Status X was correct, we can compare it with 1/n and we can also compare it with 1/(n^1/2)

both will give divergence as result by using Basic Comparison test.

Now I am having another problem which looks like to me of the same sort. I don't know its correct answer. here is the problem :

\sum \frac{1}{\sqrt{n}+\sqrt{n+1}}

we have to find that is the following series converging or diverging??

here is what i have done:

The Divergence Test seems to be failed here as it is giving me an answer equals to zero!

Now if i go for basic comparison test, what series should i consider in order to compare it with above series?

I have worked on the following series and they don't give me proper results:

\sum \frac{1}{\sqrt{n}}

\sum \frac{1}{\sqrt{n+1}}

\sum \frac{1}{n}

\sum \frac{1}{n^(1.5)}

Then what should i consider?

please answer this question as soon as possible!

thanks in advance!
 
Last edited:
What do the terms look like as n gets very large? That should give you an idea of what to compare it to.
 

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