Construct an alternating series such that $\sum (-1)^n a_n$ converge but $\sum (-1)^n a_n^3$ diverge where $a_n>0$

Question

if I write it as the follow:

$a_1-a_2+a_3-a_4...$ converge but $a_1^3-a_2^3+a_3^3-a_4^3...$ diverge

What I have thought is that by making the odd term($a_{2k+1}^3$) a diverge series like $1,\frac{1}{3}, \frac{1}{5}...$, and the even term($a_{2k}^3$) as something converge with a smaller order such as $\frac{1}{2^3},\frac{1}{4^3}...$, then $\sum (-1)^n a_n^3=1-\frac{1}{2^3}+\frac{1}{3}-\frac{1}{4^3}...$ would be a diverge series, but I can not come up with a satisfactory result to make the original series converge

Hope you could help

Answer 1

Consider the following sequence for $n \geq 0$: \begin{align*} x_{4n} = x_{4n+2} &= \frac{2}{\sqrt[3]{n+1}}\\ x_{4n+1} &= \frac{3}{\sqrt[3]{n+1}}\\ x_{4n+3} &= \frac{1}{\sqrt[3]{n+1}} \end{align*} Then $\sum (-1)^n x_n$ converges to $0$ whereas $\sum (-1)^n x_n^3$ diverges since $$x_{4n}^3 - x_{4n+1}^3 + x_{4n+2}^3 - x_{4n+3}^3 = -\frac{12}{n+1}.$$