How to do it?

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

2 Answers
Nov 10, 2016

Answer:

#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#

Nov 10, 2016

Answer:

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.