Sequence question problem

Question

Given the sequence $a_n=\sqrt[3]{n^3+c}-n$ where $c > 0,$ it's a real const. Show that $0<a_n<\frac{c}{3n^2}$ is true, for all $n$ that belong to Natural Numbers. Hint: $(n+a_n)^3=n^3+c$.

Answer 1

Hints:

• If $a>b>0,$ then $$\sqrt[3]a>\sqrt[3]b$$

• $(n+a_n)^3=n^3+3n^2a_n+3na_n^2+a_n^3=n^3+c$ thus $$a_n(3n^2+3na_n+a_n^2)=c$$

• If $x,y,k>0$, then $$\frac{x}{y}>\frac{x}{y+k}$$