<s>[INST] Does the convergence of $\sum (a_k)^2$ imply $\sum (a_k)^3$ convergence? Does the convergence of $\sum (a_k)^2$ imply $\sum (a_k)^3$ convergence? I feel like it definitely should but can't find a solid way to prove it ....
 [/INST] It's true for real sequences (as shown in other answers), but false for complex sequences.
For example, if $\omega = e^{2\pi i/3}$ and
$$
a_{3k+m} = \frac{\omega^m}{(k+1)^{1/3}}
\qquad\text{for}\qquad
k \ge 0
,\qquad
m \in \{ 0,1,2 \}
,
$$
that is, if the sequence $(a_j)_0^\infty$ is
$$
\frac{1}{1^{1/3}},
\frac{\omega}{1^{1/3}},
\frac{\omega^2}{1^{1/3}},
\quad
\frac{1}{2^{1/3}},
\frac{\omega}{2^{1/3}},
\frac{\omega^2}{2^{1/3}},
\quad
\ldots,
\quad
\frac{1}{n^{1/3}},
\frac{\omega}{n^{1/3}},
\frac{\omega^2}{n^{1/3}},
\quad
\ldots
,
$$
then $\sum a_j^2$ converges (to zero). Indeed, each group of three terms sums to zero:
$$
\left( \frac{1}{n^{1/3}} \right)^2 +
\left( \frac{\omega}{n^{1/3}} \right)^2 +
\left( \frac{\omega^2}{n^{1/3}} \right)^2 
=
\frac{1+\omega^2+\omega^4}{n^{2/3}}
=
0
,
$$
so the partial sums $S_n = \sum_{j=0}^{n-1} a_j^2$ are $S_{3k}=0$, $S_{3k+1}=a_{3k}^2$, $S_{3k+2}=a_{3k}^2+a_{3k+1}^2$,
and thus $S_n \to 0$ since $a_j^2 \to 0$.
But $\sum a_j^3$ diverges, since
$$
\left( \frac{1}{n^{1/3}} \right)^3 +
\left( \frac{\omega}{n^{1/3}} \right)^3 +
\left( \frac{\omega^2}{n^{1/3}} \right)^3 
=
\frac{1+1+1}{n}
$$
and $\sum (1/n)$ diverges.
</s>