# How to do it?

## What is the least positive integer $n$ for which ${\left(n + 1\right)}^{\frac{1}{3}} - {n}^{\frac{1}{3}} < \frac{1}{12}$ satisfies?

Nov 10, 2016

$n = 8$

#### Explanation:

Note that:

${\left(x + \frac{1}{12}\right)}^{3} = {x}^{3} + 3 \left(\frac{1}{12}\right) {x}^{2} + 3 {\left(\frac{1}{12}\right)}^{2} x + {\left(\frac{1}{12}\right)}^{3}$

$\textcolor{w h i t e}{{\left(x + \frac{1}{12}\right)}^{3}} = {x}^{3} + \frac{1}{4} {x}^{2} + \frac{1}{48} x + \frac{1}{1728}$

Given:

${\left(n + 1\right)}^{\frac{1}{3}} - {n}^{\frac{1}{3}} < \frac{1}{12}$

Add ${n}^{\frac{1}{3}}$ to both sides to get:

${\left(n + 1\right)}^{\frac{1}{3}} < {n}^{\frac{1}{3}} + \frac{1}{12}$

Cube both sides to get:

$\textcolor{red}{\cancel{\textcolor{b l a c k}{n}}} + 1 < \textcolor{red}{\cancel{\textcolor{b l a c k}{n}}} + \frac{1}{4} {n}^{\frac{2}{3}} + \frac{1}{48} {n}^{\frac{1}{3}} + \frac{1}{1728}$

Subtract $n$ from both sides and multiply both sides by $1728$ to get:

$1728 < 432 {n}^{\frac{2}{3}} + 36 {n}^{\frac{1}{3}} + 1$

Subtract $1728$ from both sides, transpose and substitute $x = {n}^{\frac{1}{3}}$ to get:

$432 {x}^{2} + 36 x - 1727 > 0$

Rather than solve this properly, let's approximate the positive zero:

$x \approx \sqrt{\frac{1727}{432}} \approx \sqrt{4} = 2$

Then $n = {x}^{3} \approx 8$

We find:

${\left(8 + 1\right)}^{\frac{1}{3}} - {8}^{\frac{1}{3}} = \sqrt{9} - 2 \approx 0.08008 < 0.08 \overline{3} = \frac{1}{12}$

Nov 10, 2016

8

#### Explanation:

Let n > 1. Then 1/n < 1.

Use ${\left(1 + x\right)}^{m} = 1 + m {C}_{1} x + m {C}_{2} {x}^{2} + \ldots , x \in \left(- 1 , 1\right]$

${\left(n + 1\right)}^{\frac{1}{3}}$

$= {n}^{\frac{1}{3}} {\left(1 + \frac{1}{n}\right)}^{- \frac{2}{3}}$

=n^(1/3)(1+(1/3)(1/n)+((1/3)(1/3-1))/(2!)(1/n^2)+.....

$= {n}^{\frac{2}{3}} + \left(\frac{1}{3}\right) {n}^{- \frac{2}{3}} - \left(\frac{1}{9}\right) {n}^{- \frac{5}{3}} + \ldots$ -

Here, the absolute values of the terms form a diminishing

sequence. It follows that

${\left(n + 1\right)}^{\frac{1}{3}} - {n}^{\frac{1}{3}} < \left(\frac{1}{3}\right) {n}^{- \frac{2}{3}} < \frac{1}{12}$

And so, ${n}^{- \frac{2}{3}} < \frac{1}{4} \to {n}^{\frac{2}{3}} > 4 \to n > {4}^{\frac{3}{2}} = 8$

As the value of the expression, for n = 7,

$= 2 - {7}^{\frac{1}{3}} = 0.087 \ldots > \frac{1}{12} = 0.0833 . .$,