If $\sum a_n$ converges does it imply $\sum a_n^3$ converges? As title says does the converges of $\sum_{i=0}^\infty a_i$ imply that $\sum_{i=0}^\infty (a_i)^3$ converge?
If we assume $a_i\leq 0$ or $a_i\geq 0$ from some point on the statement is obviously true.
If we remove the crucial assumption about the sign, it might not. Let
$$ a_n = \frac{1}{\log^2(n+2)}\cdot\left\{\begin{array}{lcl}1&amp;\text{if}&amp;n\equiv 1\pmod{3}\\1&amp;\text{if}&amp;n\equiv 2\pmod{3}\\-2&amp;\text{if}&amp;n\equiv 0\pmod{3}\end{array}\right.$$
It it simple to check that $\sum_{n\geq 0}a_n$ is convergent, but $\sum_{n\geq 0}a_n^3$ is not, essentially by Kronecker's lemma (summation by parts).
$\textbf{Hint: }$ If 
$$\sum_{n=1}^{\infty}a_n$$
converge, then $a_n\to 0$ as $n\to \infty$. Which means, there exists some $N\in \mathbb{N}$ such that $|a_n|&lt;1$ for all $n\ge N$. Then $|a_n|^3=|a_n^3|&lt;|a_n|$, and hence
$$\sum_{n=N}^\infty a_n^3$$
converge (the absolute convergence is necessary).