# Lim of An=(n+1)^(1/3) - n^(1/3)

1. Feb 10, 2012

### oferon

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: Feb 10, 2012
2. Feb 10, 2012

### kai_sikorski

$$(n+x)^{1/3} = n^{1/3}+\frac{x}{3 n^{2/3}}-\frac{x^2}{9 n^{5/3}}+\frac{5 x^3}{81 n^{8/3}} + ...$$
$$(n+1)^{1/3} - (n)^{1/3} = \frac{1}{3 n^{2/3}} + O(n^{-5/3})$$
$$\lim_{n \to \infty} \left((n+1)^{1/3} - (n)^{1/3}\right) = 0$$

3. Feb 10, 2012

### oferon

This is the binomal theorem right? Is this the only way to solve this lim?

4. Feb 10, 2012

### kai_sikorski

I just expanded using a Taylor series. There's probably other ways.

5. Feb 10, 2012

### oferon

Yep, the trick is using $$a^3-b^3=(a-b)(a^2+ab+b^2)$$

6. Feb 10, 2012

### kai_sikorski

Not sure how that helps, but if you found some way it does great!

7. Feb 10, 2012

### kai_sikorski

Oh okay I see. You know it's positive and then you bound it above using what you wrote.

8. Feb 12, 2012

### checkitagain

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

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

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

What does the numerator become?

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