Why does 1/[n log(n)]^1.1 converge

  • Thread starter grossgermany
  • Start date
In summary, convergence in this context refers to the behavior of a mathematical sequence or series as its terms approach a certain limit. The convergence of 1/[n log(n)]^1.1 is determined by analyzing its behavior as n approaches infinity using mathematical techniques such as the limit comparison test or the integral test. The exponent 1.1 in 1/[n log(n)]^1.1 represents the rate at which the terms of the sequence decrease, and the term log(n) slows down this rate, making it converge more slowly. Understanding the convergence of this sequence can have applications in various fields, such as computer science, economics, and physics, providing insights into the behavior of different systems and helping make predictions about their future behavior
  • #1
grossgermany
53
0

Homework Statement


Prove that the series:
1/[n log(n)]^1.1 converges

Homework Equations




The Attempt at a Solution




We know that nlogn is equal to d[log(log(n))] and use the integral test to show that it diverges.
However, I have no idea how to deal with the 1.1th power.
 
Physics news on Phys.org
  • #2
n > 10

log(n) >1

Thus,

1/[(nlog(n))^(1.1)] < 1/ (n^1.1)
 

FAQ: Why does 1/[n log(n)]^1.1 converge

1. What does the term "converge" mean in this context?

Convergence refers to the behavior of a mathematical sequence or series as its terms approach a certain limit. In this case, we are looking at whether the sequence 1/[n log(n)]^1.1 approaches a specific value as n (the number of terms) increases.

2. How is the convergence of 1/[n log(n)]^1.1 determined?

The convergence of a sequence or series is determined by analyzing its behavior as n approaches infinity. In this case, we can use mathematical techniques such as the limit comparison test or the integral test to determine if 1/[n log(n)]^1.1 converges or diverges.

3. What is the significance of the exponent 1.1 in 1/[n log(n)]^1.1?

The exponent 1.1 represents the rate at which the terms of the sequence decrease. In this case, the sequence decreases at a rate of 1/[n log(n)]^(1.1), which is slower than 1/n or 1/n^2. The specific exponent was chosen to make the sequence converge in a useful way.

4. How does the term "log(n)" affect the convergence of 1/[n log(n)]^1.1?

The term log(n) slows down the rate at which the sequence decreases, making it converge more slowly than if it were just 1/n^1.1. This can be seen by comparing the convergence of 1/n^1.1 and 1/[n log(n)]^1.1 as n approaches infinity.

5. What are some real-world applications of understanding the convergence of 1/[n log(n)]^1.1?

Understanding the convergence of this sequence can have applications in various fields, such as computer science, economics, and physics. In computer science, it can help in analyzing the efficiency of algorithms. In economics, it can be used to model the growth of certain populations. In physics, it can be used to study the behavior of certain physical systems. Overall, understanding the convergence of this sequence can provide insights into the behavior of various systems and help make predictions about their future behavior.

Similar threads

Back
Top