Question

How to do it?

What is the least positive integer `n` for which `(n+1)^(1/3)-n^(1/3)<1/12` satisfies?

Answer 1

`n = 8`

Explanation:

Note that:

`(x + 1/12)^3 = x^3 + 3(1/12)x^2 + 3(1/12)^2x+(1/12)^3`

`color(white)((x + 1/12)^3) = x^3 + 1/4x^2 + 1/48x+1/1728`

Given:

`(n+1)^(1/3)-n^(1/3) < 1/12`

Add `n^(1/3)` to both sides to get:

`(n+1)^(1/3) < n^(1/3)+1/12`

Cube both sides to get:

`color(red)(cancel(color(black)(n)))+1 < color(red)(cancel(color(black)(n)))+1/4 n^(2/3) + 1/48 n^(1/3) + 1/1728`

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

`1728 < 432 n^(2/3) + 36 n^(1/3) + 1`

Subtract `1728` from both sides, transpose and substitute `x = n^(1/3)` to get:

`432 x^2 + 36 x - 1727 > 0`

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

`x ~~ sqrt(1727/432) ~~ sqrt(4) = 2`

Then `n = x^3 ~~ 8`

We find:

`(8+1)^(1/3) - 8^(1/3) = root(3)(9) - 2 ~~ 0.08008 < 0.08bar(3) = 1/12`

Answer 2

8

Explanation:

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

Use `(1+x)^m=1+mC_1x+mC_2x^2+..., x in (-1, 1]`

`(n+1)^(1/3)`

`=n^(1/3)(1+1/n)^(-2/3)`

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

`=n^(2/3)+(1/3)n^(-2/3)-(1/9)n^(-5/3)+...` -

Here, the absolute values of the terms form a diminishing

sequence. It follows that

`(n+1)^(1/3)-n^(1/3)<(1/3)n^(-2/3)< 1/12`

And so, `n^(-2/3)<1/4 to n^(2/3)>4 to n > 4^(3/2)=8`

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

`=2-7^(1/3)=0.087...>1/12=0.0833..`,

the answer is 8.