How Do You Simplify the Limit of (n+1)^(1/3) - n^(1/3)?

  • Context: Undergrad 
  • Thread starter Thread starter oferon
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around the limit of the expression \(\lim_{n\to\infty}\sqrt[3]{n+1} - \sqrt[3]{n}\). Participants explore various methods to simplify this limit, including the use of series expansions and algebraic identities.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that the limit is 0 but seeks clarification on how to simplify the expression to demonstrate this.
  • Another participant provides a series expansion for \((n+x)^{1/3}\) and concludes that the limit approaches 0, presenting a specific form of the expression.
  • A participant questions whether the binomial theorem is the only method to solve the limit.
  • Another mentions using a Taylor series as an alternative approach.
  • One participant introduces the identity \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\) as a potential method for simplification.
  • There is a discussion about bounding the expression and the implications of the denominator as \(n\) approaches infinity.

Areas of Agreement / Disagreement

Participants express various methods for approaching the limit, but there is no consensus on a single preferred method or whether the binomial theorem is the only approach. The discussion remains unresolved regarding the best technique to simplify the limit.

Contextual Notes

Some participants rely on specific mathematical identities and series expansions, while others express uncertainty about the applicability of these methods. The discussion includes assumptions about the behavior of the expressions as \(n\) approaches infinity.

oferon
Messages
29
Reaction score
0
Hi all!
Been trying to look for some examples with no luck.. all I found is
related to square roots, not cube roots..
Anyway I'm trying to solve: \lim_{n\to\infty}\sqrt[3]{n+1} - \sqrt[3]{n}

The limit is obviously 0.. But how do I simplify this expression to show it?
Or should I use the ratio\root test? Even though I couldn't..
Thanks a bunch
 
Last edited:
Physics news on Phys.org
(n+x)^{1/3} = n^{1/3}+\frac{x}{3 n^{2/3}}-\frac{x^2}{9 n^{5/3}}+\frac{5 x^3}{81 n^{8/3}} + ...
(n+1)^{1/3} - (n)^{1/3} = \frac{1}{3 n^{2/3}} + O(n^{-5/3})
\lim_{n \to \infty} \left((n+1)^{1/3} - (n)^{1/3}\right) = 0
 
This is the binomal theorem right? Is this the only way to solve this lim?
 
I just expanded using a Taylor series. There's probably other ways.
 
Yep, the trick is using a^3-b^3=(a-b)(a^2+ab+b^2)

Thanks for your help anyway.
 
Not sure how that helps, but if you found some way it does great!
 
Oh okay I see. You know it's positive and then you bound it above using what you wrote.
 
oferon said:
Yep, the trick is using a^3-b^3=(a-b)(a^2+ab+b^2)

Thanks for your help anyway.

Let \ a \ = \ \ \sqrt[3]{n + 1} \ \ and \ \ b \ \ = \ \sqrt[3]{n}.


Then \ \ \dfrac{a - b}{1} \cdot \dfrac{a^2 + ab + b^2}{a^2 + ab + b^2} \ =


\dfrac{a^3 - b^3}{a^2 + ab + b^3}



What does the numerator become?


As n --> oo, what happens to the denominator?
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
2
Views
1K