Bounds of a sequence of numbers

Question

If we define a sequence $a_n$ where $n\in\mathbb{N}$ by $$a_n=\frac{1}{({2n\pi})^{\frac{1}{3}}}$$ Then how could one show the inequality $$a_n-a_{n+1}\leq Cn^{-\frac{4}{3}}$$

I have tried got to a point where $$a_n-a_{n+1}\leq(\frac{2\pi}{(2n\pi)(2(n+1)\pi)})^{1/3}$$

Answer 1

We have that

$$a_n-a_{n+1}=\frac{1}{({2n\pi})^{\frac{1}{3}}}-\frac{1}{( {2(n+1)\pi})^{\frac{1}{3}} }=\frac{1}{{(2(n+1)\pi})^{\frac{1}{3}}}\left(\left(1+\frac1n\right)^\frac13-1\right)\le$$

$$\le\frac{1}{(2\pi)^\frac13n^\frac13}\left(1+\frac1{3n}-1\right)=\frac{1}{3(2\pi)^\frac13}\frac1{n^\frac43}$$

indeed

• by Bernoulli inequality $\left(1+\frac1n\right)^\frac13\le 1+\frac1{3n}$
• $\frac1{(n+1)^\frac13}\le \frac1{n^\frac13}$

Answer 2

We can factor out the $(2\pi)^{-1/3}$ in each term. We're left with $n^{-1/3}-(n+1)^{-1/3}.$ By the MVT, this equals

$$(-1/3)c^{-4/3}\cdot (-1), \, \text { for some }c\in (n,n+1).$$

This is less than $(1/3)n^{-4/3}$ and we're done.