$\sum^{\infty}_{n=1}x_n$ converges $\Rightarrow$ $\sum^{\infty}_{n=1}x_n^3$ converges also? Let the series $\sum^{\infty}_{n=1}x_n$ converge. $\{x_n\}_{n=1}^{\infty} \in \mathbb R$
  
  Does $\sum^{\infty}_{n=1}x_n^3$ converge too?


I tried to find some counter-examples but found none. I tried to prove this also, but I can't... I don't know even if it's true or not.
Let $a= \exp(2 \pi i/3)$ (such that $1+a+a^2=0$ and $a^3=a^6=1$)

Take the series $$x_{3n} = \frac{1}{n^{1/3}},\quad x_{3n+1} = ax_{3n},\quad x_{3n+2}=a^2x_{3n}.$$

With such choice of $x$, the series $\sum_n x_n$ converges, but the series $\sum_n x_n^3$ does not.

edit

Choose real numbers $a$, $b$, $c$ such that $$a+b+c=0\\a^3+b^3+c^3&gt;0.$$
For example, $a=-1$, $b=-2$, $c=3$. Then take 

$$x_{3n} = \frac{a}{n^{1/3}},\quad x_{3n+1} = \frac{b}{n^{1/3}},\quad x_{3n+2}=\frac{c}{n^{1/3}}.$$

With such choice of $x$, the series $\sum_n x_n$ converges, but the series $\sum_n x_n^3$ does not.
Hint : justify that there exists $N$ such that for $n\geq N$ :

$$(x_n)^3\leq x_n $$

and conclude.