# Question

Find the least possible no for which $$\sqrt[3]{n+1} - \sqrt[3] n< \frac 1{12}$$

How do I reach a stage where I can deduce definitely the smallest integer value of $n$. I keep getting stuck after I cube on both sides and I still get cube root terms which I can't seem to simplify . Any help is appreciated.Thanks :)

• What exactly in the point in that $1$ as exponent??? – barak manos Dec 22 '16 at 13:02
• Use {} to enclose the exponents so they stay up in the air :) – Erick Wong Dec 22 '16 at 13:02
• Are you sure that the question says "smallest" and not "largest" (or "$<$" and not "$>$")??? Any integer $n<1$ satisfies this condition, so there is no minimum such bound for $n$. – barak manos Dec 22 '16 at 13:03
• Not sure what you mean by the "least possible" $n$. Of course, you don't specify any range for $n$, but large negative numbers work. Did you mean "greatest possible"? – lulu Dec 22 '16 at 13:04
• @barak manos I just checked and it says least possible integer – Shash Dec 22 '16 at 13:05

Let $n = x^3$

\begin{array}{c} \sqrt[3]{n+1} - \sqrt[3] n < \frac 1{12} \\ \sqrt[3]{x^3+1} - x < \frac 1{12} \\ \sqrt[3]{x^3+1} < x + \frac 1{12} \\ x^3+1 < x^3 + \dfrac 14x^2 + \dfrac{1}{48}x + \dfrac{1}{1728} \\ \dfrac 14x^2 + \dfrac{1}{48}x - \dfrac{1727}{1728} > 0 \\ x^2 + \dfrac{1}{12}x - \dfrac{1727}{432} > 0 \\ \left( x + \dfrac{1}{24} \right)^2 - \dfrac{6911}{1728} > 0\\ x + \dfrac{1}{24} > \dfrac{\sqrt{6911}}{24\sqrt 3} \\ x > \dfrac{\sqrt{6911}}{24\sqrt 3} - \dfrac{1}{24} \\ x > \dfrac{1}{72} (\sqrt{20733} - 3) \\ x^3 > \left(\dfrac{1}{72} (\sqrt{20733} - 3)\right)^3 \\ n > \dfrac{865 \sqrt{20733} - 7776}{15552} \\ n > 7.5 \end{array} The smallest value is $8$.

Since the function is convex, you have that: $$f(x+dx) < f(x) + f'(x)dx$$ $$dy <f'(x)dx$$ In this case we have: $$dy =\sqrt[3]{n+1}- \sqrt[3]{n} < \frac{1}{3}n^{-2/3}$$ We want the first $n$ such that $dy<1/12$. If we compare both values we get that if (but not only if!) $n>8$, your condition holds. It is enough to check that $n=7$ does not satisfy the condition to find $n=8$.

HINT:

$$\sqrt[3]{n+1}-\sqrt[3] n=\dfrac{n+1-n}{(n+1)^{2/3}+(n+1)^{1/3}n^{1/3}+n^{2/3}}$$

Now $3n^{2/3}<(n+1)^{2/3}+(n+1)^{1/3}n^{1/3}+n^{2/3}<3(n+1)^{2/3}$

Here's a simple way to get an upper bound on the least $n$ without doing much in the way of algebra or numerical calculations:

Note that the average value of the first $26$ values of $\sqrt[3]{n+1}-\sqrt[3]n$ is a telescoping sum,

$${(\sqrt[3]{27}-\sqrt[3]{26})+(\sqrt[3]{26}-\sqrt[3]{25})+\cdots+(\sqrt[3]{3}-\sqrt[3]{2})+(\sqrt[3]{2}-\sqrt[3]{1})\over26}={3-1\over26}={1\over13}$$

Since ${1\over13}\lt{1\over12}$, at least one of the numbers in the average must be less than $1\over12$. So the number being sought is no greater than $26$.

This only tells you where to look, of course. If there is a simple way to zero in on the exact answer that doesn't use some combination of algebra and numerics along the lines of what's been done in other answers, I'd like to see it.