True or false:
If $\sum\limits_{n = 1}^\infty {{a_n}} $ is a convergent series of real numbers then $\sum\limits_{n = 1}^\infty {{{\left( {{a_n}} \right)}^3}} $ is also convergent.
My attempt:
Note that since the series converges, we have $\mathop {\lim }\limits_{n \to \infty } {a_n} = 0$ and thus ${a_n} \le 1$ for a large enough $n$.
The claim is obviously true if ${a_n} \ge 0$ since if that's the case then ${\left( {{a_n}} \right)^3} \le {a_n}$ for a large enough $n$ and by a comparison theorem (for non negative series) we conclude that $\sum\limits_{n = 1}^\infty {{{\left( {{a_n}} \right)}^3}} $ converges.
The claim is also true if ${a_n} < 0$ for all natural $n$ by symmetry. Since ${a_n} < 0$ iff ${\left( {{a_n}} \right)^3} < 0$.
But I'm stuck if ${a_n}$ is sometimes positive and sometimes negative. I've tried to take a "classic" example like $$\sum\limits_{n = 1}^\infty {{{{{\left( { - 1} \right)}^n}} \over n}} $$ But that doesn't disprove the claim.
Hints would be welcome!
