Sqrt(4n)/sqrt(4n-3)sqrt(4n^2-3n) diverges?

  • Thread starter grossgermany
  • Start date
In summary, to show that sqrt(4n)/sqrt(4n-3)sqrt(4n^2-3n) diverges, it is sufficient to use the limit comparison test with the harmonic series. However, for a rigorous proof in a real analysis class, one would need to show that the limit an/bn is greater than or equal to 1 for all n greater than or equal to some value n0.
  • #1
grossgermany
53
0

Homework Statement


How to show that sqrt(4n)/sqrt(4n-3)sqrt(4n^2-3n) diverges?


Homework Equations





The Attempt at a Solution


The above expression is asymptotically equivalent to 1/2n which diverges as the harmonic series diverges.

However, a rigorous proof is required for the real analysis class.
 
Physics news on Phys.org
  • #2
grossgermany said:

Homework Statement


How to show that sqrt(4n)/sqrt(4n-3)sqrt(4n^2-3n) diverges?


Homework Equations





The Attempt at a Solution


The above expression is asymptotically equivalent to 1/2n which diverges as the harmonic series diverges.

However, a rigorous proof is required for the real analysis class.
How rigorous? Would the limit comparison test be rigorous enough?
 
  • #3
Yes, comparison test would be great.
 
  • #4
I would use the limit comparison test, not the comparison test. For the comparison test, and some series [itex]\sum b_n[/itex] that is known to diverge, you would have to show that an >= bn for all n >= n0.

For the limit comparison test, you only need to look at lim an/bn.
 

Related to Sqrt(4n)/sqrt(4n-3)sqrt(4n^2-3n) diverges?

1. Why does the expression Sqrt(4n)/sqrt(4n-3)sqrt(4n^2-3n) diverge?

The expression diverges because as n approaches infinity, the numerator and denominator both approach infinity, resulting in an infinite value. This is known as an infinite limit and indicates that the expression does not have a finite value.

2. Can the expression Sqrt(4n)/sqrt(4n-3)sqrt(4n^2-3n) be simplified?

Yes, the expression can be simplified by factoring out a common term of sqrt(4n) from both the numerator and denominator. This results in the simplified expression of 1/sqrt(1-3/n).

3. How is the divergence of Sqrt(4n)/sqrt(4n-3)sqrt(4n^2-3n) related to the concept of infinity?

The divergence of the expression is related to the concept of infinity because as n approaches infinity, the expression becomes unbounded and has no finite value. This is a characteristic of infinite limits.

4. Is there a way to make the expression Sqrt(4n)/sqrt(4n-3)sqrt(4n^2-3n) converge?

No, there is no way to make the expression converge as it is defined. However, if we restrict the values of n to a finite interval, the expression may converge. This is known as a bounded interval.

5. How is the divergence of Sqrt(4n)/sqrt(4n-3)sqrt(4n^2-3n) relevant in real-life applications?

The divergence of the expression is relevant in real-life applications because it represents a situation where a quantity becomes unbounded and does not have a finite value. This can occur in various scenarios such as population growth or resource depletion.

Similar threads

  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
142
  • Calculus and Beyond Homework Help
Replies
5
Views
934
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
3K
Replies
2
Views
1K
Back
Top