If $\sum a_n$ converges then $\sum a_n^3$ converges Prove or contradict: if $\sum a_n$ converges then $\sum a_n^3$ converges.
I was able to prove that if $a_n \geq 0$ then the statement is true. But I couldn't prove nor contradict the general case.

False, counterexample: $$a_n = \frac{\epsilon_n}{\sqrt[3]{\lceil n/3 \rceil}}
\quad\text{ where }\quad \epsilon_n = \begin{cases}+2,& n \equiv 1, \pmod 3\\
-1, & n \not\equiv 1, \pmod 3\end{cases}$$
It is easy to see 
$\displaystyle\;\left| \sum_{n=1}^N a_n \right| \le \frac{2}{\sqrt[3]{\lceil N/3 \rceil}}
\quad\implies\quad
\sum_{n=1}^\infty a_n$ exists and equal to $0$.
However $\displaystyle\;\sum_{n=1}^{3N} a_n^3 = 6\sum_{n=1}^N \frac{1}{n} \approx 6( \log N + \gamma )\;$
diverges logarithmically.
