Will This Infinite Series Converge or Diverge?

  • Thread starter Monte_Cristo
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In summary: The integral test is not a good test to use on this problem because the series is infinite. The series 1/(1+n*ln(n)^2) does not converge, but it also does not diverge. So it seems like the infinite series diverges.
  • #1
Monte_Cristo
3
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Homework Statement



Determine whether the infinite series converge or diverge?


∑ 1/{1+n[(ln n)^2]}
n=0



Homework Equations



Direct Comparison test
Limit Comparison test
Root test
Ratio test
integral test



The Attempt at a Solution



Personally, I think that only two methods can work:

1) Manual method - meaning writing down the series and then look at if its converging or diverging.

2) Direct Comparison test - If I compare it directly with the limit of n approaching infinity for 1/n, then my answer comes out to be that the "series diverges

If I use 1/n[(ln n)^2] or else 1/[(ln n)^2] for direct comparison test, I get the same result.

BUT, when I do it the manual way, my answer turns out to be CONVERGENT.

Its so confusing, please help!

Thanks
 
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  • #2
Monte_Cristo said:

Homework Statement



Determine whether the infinite series converge or diverge?


∑ 1/{1+n[(ln n)^2]}
n=0



Homework Equations



Direct Comparison test
Limit Comparison test
Root test
Ratio test
integral test



The Attempt at a Solution



Personally, I think that only two methods can work:

1) Manual method - meaning writing down the series and then look at if its converging or diverging.
I have no idea of what you mean by the "manual" method.
Monte_Cristo said:
2) Direct Comparison test - If I compare it directly with the limit of n approaching infinity for 1/n, then my answer comes out to be that the "series diverges
Maybe or maybe not. Did you show that each term of your series is larger than the corresponding term of the series [itex]\sum 1/n[/itex]?
Monte_Cristo said:
If I use 1/n[(ln n)^2] or else 1/[(ln n)^2] for direct comparison test, I get the same result.
Do you know for a fact that these series diverge?
Monte_Cristo said:
BUT, when I do it the manual way, my answer turns out to be CONVERGENT.

Its so confusing, please help!

Thanks
 
  • #3
oh my goodness! Now I see that why the Direct Comparison Test can not work on this solution.

By saying "manual way", I mean that if I write down the series (a1 + a2 + a3...an), then my answer turns out to be convergent.

Thanks for your help! But it would be good if I can get some more! :) This problem is hard! All I know is that I am NOT supposed to solve this problem in the "Manual way" - I need to use one of the tests to prove it - and I am TOTALLY stuck! :)

Thanks
 
  • #4
But your series is an infinite series, so there is no way you can write all of its terms. They way you have written it -(a1 + a2 + a3...an) - suggests that there is a last term, which is not the case for infinite series.

You listed several tests that you can use. I wouldn't try the integral test or the n-th root test, but I would try the limit comparison and ratio tests, and maybe the direct comparison test. When you use the two comparison tests, you have to have a few series in your pocket to compare against, such as sum(1/n) and sum(1/n^2), two series that respectively diverge and converge.
 
  • #5
Are you sure the summation is from 0 to [tex]\infty[/tex], because ln(n) doesn't exist for n=0. Maybe it's supposed to be a trick question?
 
  • #6
Atropos said:
Are you sure the summation is from 0 to [tex]\infty[/tex], because ln(n) doesn't exist for n=0. Maybe it's supposed to be a trick question?

I deeply apologize, the summation is:


∑ 1/{1+n[(ln n)^2]}
n=1

I am so sorry that I made the typo error. lol but still I can't find the solution! :)
 
  • #7
Apply the integral test to 1/(n*ln(n)^2). What do you conclude? How does that series compare with 1/(1+n*ln(n)^2)?
 

FAQ: Will This Infinite Series Converge or Diverge?

1. What is the difference between convergence and divergence?

Convergence refers to the behavior of a sequence or series as its terms approach a fixed value or limit. Divergence, on the other hand, indicates that the terms of the sequence or series do not approach a fixed value and may instead grow infinitely large.

2. How can I determine if a series converges or diverges?

There are a variety of tests that can be used to determine the convergence or divergence of a series, including the comparison test, ratio test, and integral test. These tests involve comparing the series to known convergent or divergent series or evaluating the limit of a function.

3. What does it mean when a series converges?

When a series converges, it means that the sum of its terms approaches a finite value as the number of terms increases. In other words, the series has a finite sum or limit.

4. Can a series both converge and diverge?

No, a series cannot both converge and diverge. It will either approach a finite limit (converge) or grow infinitely large (diverge).

5. What is the importance of determining convergence or divergence?

Determining the convergence or divergence of a series is important in many areas of mathematics and science. It allows us to evaluate the behavior of a sequence or series, make predictions about its future terms, and determine the validity of mathematical statements and proofs.

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