# How to do it?

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

What is the least positive integer

##### 2 Answers

#### Answer:

#### 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)^(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

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

Subtract

#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

We find:

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

#### Answer:

8

#### Explanation:

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

Use

Here, the absolute values of the terms form a diminishing

sequence. It follows that

And so,

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

the answer is 8.