$\sum^{\infty}_{n=1}x_n$ 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.
 A: 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.
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Choose real numbers $a$, $b$, $c$ such that $$a+b+c=0\\a^3+b^3+c^3>0.$$
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.
A: No, if the signs of the $x_n$ are not the same; here's a fairly typical counterexample.  Define the series $\sum_nx_n$ as follows. Its positive terms are $1/n^{1/3}$ for all natural numbers $n$, each of these terms occurring exactly once. Its negative terms are $-1/n^{4/3}$, each occurring exactly $n$ times, immediately after the corresponding positive term $1/n^{1/3}$.  In other words, the series consists of blocks of consecutive terms, where the $n$th black consists of $1/n^{1/3}$ followed by $n$ consecutive occurrences of $-1/n^{4/3}$. Notice that each block has sum $0$, from which it easily follows that the whole series converges to $0$.  Now consider what happens when you cube each term. The $n$th block of the new series consists of a positive term $1/n$ followed by $n$ consecutive negative terms $-1/n^4$.  The $n$ negative terms in the $n$th block add up to $-1/n^3$ which is far from cancelling the positive term $1/n$.  Indeed, except for the first block or two, the sum of the $n$th block will be positive and larger than $1/(2n)$.  Since the harmonic series diverges, this means that the series of cubes diverges to $+\infty$.
A: Hint : justify that there exists $N$ such that for $n\geq N$ :
$$(x_n)^3\leq x_n $$
and conclude.
