Does 1/n(log(n))^2 converge or diverge

  • Thread starter grossgermany
  • Start date
In summary, the conversation is discussing whether the series 1/[n(log(n))^2] converges or diverges. The participants suggest using the integral test, comparison test, ratio test, and limit comparison test, but all of these tests give inconclusive results. One participant suggests using the Exponential Integral function to evaluate the integral, but another participant points out that the bounds of the integral are unknown. Finally, it is suggested to use the method of substitution to evaluate the integral, which leads to the conclusion that the series diverges.
  • #1
grossgermany
53
0

Homework Statement



Does 1/[n(log(n))^2] converge or diverge

Homework Equations



We know that Does 1/[n(log(n))] diverges by integral test

The Attempt at a Solution

 
Physics news on Phys.org
  • #2
Then the comparison test might be useful, as long as you can show that 1/[n(log(n))^2] < 1/[n(log(n))].
 
  • #3
No sir, 1/[n(log(n))] diverges, comparison test would not help in this case.
 
  • #4
Sorry, misread what you wrote, which was clear. Does the ratio test help you? If not, try the limit comparison test.
 
  • #5
Most unfortunately both ratio test and limit comparison test give you 1 which is inconclusive.
 
  • #6
What's wrong with the integral test?
 
  • #7
EDIT: Never mind, I think I misread the question.

Just to make sure, it's x log^2(x), not (x log(x))^2, right?
 
  • #8
Char. Limit said:
Have you tried integrating [tex]\int \frac{dx}{x^2 log^2(x)}[/tex]?

It requires the Exponential Integral function Ei(u) to even be possible.

From the way he wrote it, it looks like it should be

[tex]
\frac{1}{n (log(n))^2}
[/tex]

But if you are right that he meant x^2 log^2(x) then it is trivial.

Hint: log^2(x) > 1 for x > 3.
 
  • #9
Unfortunately I have never tried the Ei(u) thing. Nor have I heard of integral function.
In other words 1/[n(log(n))^2] diverges?
 
  • #10
Yes, the integral test, however please tell me how to evaluate this integral? I suspect the result would be some expression that goes to infinity.
 
  • #11
Wait, before we continue.

We need to know your bounds.

Is the initial n n=1?

Or is the initial n n=2?
 
  • #12
How to evaluate
[tex]
\int_{2}^{\infty} \frac{1}{x log^2 (x)} dx
[/tex]

that?

Think back to Calc II. Try a u-substitution.
 
  • #13
You can also use a "bare-hands" argument, without the integral test. Imagine that the natural logarithm were instead a binary logarithm (it's just a constant factor different), and estimate the sequence by blocks whose boundaries are powers of 2.
 
  • #14
initial n=2
 
  • #15
grossgermany said:
initial n=2

Then an integral test will work just fine. Just do as l'Hopital suggested.
 
  • #16
and therefore it diverges?
 
  • #17
Use integral test and method of substitution.
 

Related to Does 1/n(log(n))^2 converge or diverge

1. What is the definition of convergence and divergence in mathematics?

Convergence and divergence refer to the behavior of a sequence or series of numbers. A sequence or series is said to converge if its terms approach a finite limit as the number of terms increases. On the other hand, if the terms of a sequence or series do not approach a finite limit, it is said to diverge.

2. Can you provide an example of a convergent series?

One example is the geometric series, where the terms decrease or increase by a constant ratio. For instance, the series 1/2 + 1/4 + 1/8 + 1/16 + ... converges to a limit of 1.

3. How does the value of n affect the convergence or divergence of 1/n(log(n))^2?

The value of n plays a crucial role in determining the convergence or divergence of 1/n(log(n))^2. As n increases, the value of the series decreases, therefore approaching a finite limit of 0. This indicates that the series converges for all values of n greater than 1.

4. Is there a specific approach to determine the convergence or divergence of a series?

Yes, there is a set of tests and criteria, such as the ratio test, integral test, and comparison test, that can be used to determine the convergence or divergence of a series. These tests involve analyzing the behavior of the series' terms and comparing them to known series with known convergence or divergence properties.

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

Determining the convergence or divergence of a series is crucial in mathematical analysis and applications. It allows us to identify the behavior of a sequence or series and predict its future values. It also helps in evaluating the accuracy of numerical methods and in making decisions in various fields such as economics, engineering, and physics.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
394
  • Calculus and Beyond Homework Help
Replies
6
Views
247
  • Calculus and Beyond Homework Help
Replies
4
Views
288
  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
838
  • Calculus and Beyond Homework Help
Replies
2
Views
806
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
328
  • Calculus and Beyond Homework Help
Replies
4
Views
979
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
Back
Top