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 ....
 A: Since $\sum (a_k)^2$ converges, $a_k \to 0$, hence ther is $K$ such that $|a_k| \le 1$ for all $k>K$.
For $k>K$ we then have $|a_k^3| \le a_k^2$. The comparison test gives that  $\sum (a_k)^3$ converges absolutely.
A: 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.
A: Hint:  Convergent sequences must have the terms have a limit of zero.   Use that and the comparison test...at what point do numbers get smaller if you cube them then if you square them?
A: Yes it does. If $\sum a_k^2$ converges, then in particular you get $a_k\to 0,$ and hence $|a_k^3|\leq a_k^2$ when $k$ is large enough(so that $|a_k|<1$). Hence, there exists $k_0$ such that $k\geq k_0$ implies $|a_k^3|\leq a_k^2.$ But then you can find that 
$$\sum |a_k^3|= \sum_1^{k_0}|a_k^3| + \sum_{k_0+1}^\infty|a_k^3|\leq \sum_1^{k_0}|a_k^3| + \sum_{k_0+1}^\infty|a_k^2|<+\infty. $$ This proves that $\sum |a_k^3|$ converges and so it does $\sum a_k^3.$
