Show that $(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$ for all positive integers

Question

I am trying to prove $$(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$$ for all positive integers.

My attempts so far have been to Taylor expand the left hand side: $$(n+1)^{2/3} -n^{2/3}\\ =n^{2/3}\big((1+1/n)^{2/3} -1\big)\\ =n^{2/3}\left(\sum_{\alpha=0}^\infty\frac{\frac{2}{3}(\frac{2}{3}-1)\cdots(\frac{2}{3}-\alpha+1)}{\alpha!} n^\alpha-1\right)\\ =n^{2/3}\sum_{\alpha=1}^\infty\frac{\frac{2}{3}(\frac{2}{3}-1)\cdots(\frac{2}{3}-\alpha+1)}{\alpha!} n^\alpha$$

I also tried proof by induction. Assume that it's true for n=k, so that $$(n+1)^{2/3} -n^{2/3} < \frac{2}{3}n^{-1/3}\\ n^{2/3}\big((1+1/n)^{2/3} -1\big)<\frac{2}{3} n^{-1/3}\\ (1+1/n)^{2/3} -1<\frac{2}{3} n^{-1}$$

Then I want to prove that $(n+2)^{2/3} -(n+1)^{2/3} < \frac{2}{3}(n+1)^{-1/3}$. The left hand side is:

$$(n+2)^{2/3} -(n+1)^{2/3}\\ =(n+1)^{2/3}\left[\left(1+\frac{1}{n+1}\right)^{2/3}-1^{2/3}\right]\\ <(n+1)^{2/3}\cdot \frac{2}{3} n^{-1}\\ =\frac{2}{3}\frac{(n+1)^{2/3}}{n^{-1}}$$ But this is bigger than $\frac{2}{3}(n+1)^{-1/3}$, so I am stumped!

Answer 1

Try multiplying both sides by $n^{1/3}$, expanding $(n(n+1)^2)^{1/3}$ to $(n^3+2n^2+n)^{1/3}$, and moving $n$ from the left hand side to the right hand side, which rewrites the inequality as

$$(n^3+2n^2+n)^{1/3}\lt n+{2\over3}$$

Cubing both sides (which is OK since $x^3\lt y^3\iff x\lt y$) turns the inequality to prove into

$$n^3+2n^2+n\lt n^3+2n^2+{4\over3}n+{8\over27}$$

which is obviously true.

Answer 2

For $f(x) = x^{\frac{2}{3}}$, then by the Mean Value Theorem,

$(n+1)^{\frac{2}{3}}-n^{\frac{2}{3}}=\frac{f(n+1)-f(n)}{(n+1)-n} = f'(c) = \frac{2}{3}c^{-\frac{1}{3}}$ for some $c \in (n,n+1)$.

Since $f'(x)=\frac{2}{3}x^{-\frac{1}{3}}=\frac{2}{3(^3\sqrt{x})}$ is decreasing, then $c > n \implies f'(c) < f'(n) = \frac{2}{3}n^{-\frac{1}{3}}$.

Answer 3

An elementary way. Let $x=n^{1/3}\geq 1$ then it suffices to show that $$(x^3+1)^{2/3} -x^2 <\frac{2}{3x}$$ that is $$(x^3+1)^2<\left(x^2+\frac{2}{3x}\right)^3$$ or $$x^6+2x^3+1<x^6+2x^3+\frac{4}{3}+\frac{8}{27x^3}$$ which trivially holds.

Answer 4

If you are allowed to integrate, then, for $a > 0$,

$\begin{array}\\ \int_n^{n+1} x^{-a} dx &= \dfrac{x^{-a+1}}{-a+1}|_n^{n+1}\\ &= \dfrac{(n+1)^{-a+1}-n^{-a+1}}{-a+1}\\ \text{and}\\ \int_n^{n+1} x^{-a} dx &\lt n^{-a} \qquad\text{since } x^{-a} \text{ is decreasing}\\ \end{array} $

Putting $a = \frac13$. this becomes $n^{-1/3} \gt \dfrac{(n+1)^{2/3}-n^{2/3}}{2/3} = \dfrac32((n+1)^{2/3}-n^{2/3}) $.

Note that, since $\int_n^{n+1} x^{-a} dx \gt (n+1)^{-a} $ we get $(n+1)^{-1/3} \lt \dfrac32((n+1)^{2/3}-n^{2/3}) $.

Answer 5

It seems nobody can resist this one :) $$\begin{align}(n+1)^{2/3}-n^{2/3}&=\left((n+1)^{1/3}-n^{1/3}\right)\left((n+1)^{1/3}+n^{1/3}\right)\\ &=\frac{\left(n+1-n\right)\left((n+1)^{1/3}+n^{1/3}\right)}{(n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}}\\ &=\frac1{n^{1/3}\left(\frac{(n+1)^{2/3}}{n^{1/3}(n+1)^{1/3}+n^{2/3}}+1\right)}\\ &=\frac1{n^{1/3}\left(\frac{\left(\frac{n+1}{n}\right)^{1/3}}{1+\left(\frac{n}{n+1}\right)^{1/3}}+1\right)}\end{align}$$ And there you have it: $\frac n{n+1}<1$, so $1+\left(\frac n{n+1}\right)^{1/3}<2$ and $$\frac{\left(\frac{n+1}{n}\right)^{1/3}}{1+\left(\frac{n}{n+1}\right)^{1/3}}+1>\frac{\left(\frac{n+1}n\right)^{1/3}}2+1>\frac12+1=\frac32$$ So $$(n+1)^{2/3}-n^{2/3}<\frac1{\frac32n^{1/3}}$$