Is the series covergent or divergent?

In summary, the conversation is about determining if a given series is convergent or divergent. The series in question is \sum \frac{2}{\sqrt{n}+1} and the person is unsure of how to solve it. They tried the Divergence Test and it gave them a result of 0, which they believed to be incorrect. They then ask for help and are advised to use the Limit Comparison Test with \sum 1/n as Bn. They also mention another series, \sum \frac{1}{\sqrt{n}+\sqrt{n+1}}, which they are unsure of how to solve and ask for help comparing it to other series. The conversation ends with a suggestion to look at the terms
  • #1
shaiqbashir
106
0
is the series covergent or divergent?

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

[tex]\sum \frac{2}{\sqrt{n}+1}[/tex]


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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  • #2
Compare it to [itex]\sum 1/n[/itex].
 
  • #3
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?
 
  • #4
Thanks for ur Help StatusX!

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

thanks a lot once again!
 
  • #5
But you know that the harmonic series (1/n) diverges, right?
 
  • #6
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 :

[tex]\sum \frac{1}{\sqrt{n}+\sqrt{n+1}}[/tex]

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:

[tex]\sum \frac{1}{\sqrt{n}}[/tex]

[tex]\sum \frac{1}{\sqrt{n+1}}[/tex]

[tex]\sum \frac{1}{n}[/tex]

[tex]\sum \frac{1}{n^(1.5)}[/tex]

Then what should i consider?

please answer this question as soon as possible!

thanks in advance!
 
Last edited:
  • #7
What do the terms look like as n gets very large? That should give you an idea of what to compare it to.
 

1. What does it mean for a series to be convergent or divergent?

A series is said to be convergent if the sum of its terms approaches a finite number as the number of terms increases. On the other hand, a series is said to be divergent if the sum of its terms approaches infinity as the number of terms increases.

2. How can I determine if a series is convergent or divergent?

There are several tests that can be used to determine if a series is convergent or divergent, such as the comparison test, the ratio test, and the integral test. These tests involve analyzing the behavior of the terms in the series and using mathematical techniques to determine if the series will approach a finite number or infinity.

3. Why is it important to know if a series is convergent or divergent?

Knowing if a series is convergent or divergent is important because it allows us to determine the behavior of the series and make predictions about its sum. This information is useful in many areas of science, including physics, chemistry, and engineering, where series are often used to model real-world phenomena.

4. Can a series be both convergent and divergent?

No, a series can only be either convergent or divergent. If the sum of the terms approaches a finite number, the series is convergent, and if the sum approaches infinity, the series is divergent. It is not possible for a series to have both of these behaviors.

5. Is there a way to transform a divergent series into a convergent one?

Yes, there are techniques such as reordering and regrouping the terms of a series that can sometimes transform a divergent series into a convergent one. However, these techniques are not always reliable and may not work for all series. It is important to carefully analyze the series and use proper mathematical techniques to determine its convergence or divergence.

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