If $\sum a_n$ converges, prove $\sum a_n^3$ converges

I can prove this if $a_n \geq 0$ for all $n$ (by comparison test). How should one tackle this in general?

Moreover, is it true that the converse also holds?

Solution 1:

That $\sum a_n$ converges does not imply $\sum a_n^3$ converges in general for non positive $a_n$.

For an $m$, write as $m=3n+k$ where $0 \leq k < 3$ and define $a_{3n+k} = b_k / (n+1)^{1/3}$ where $b_0 = 2$ and $b_1, b_2 = -1$. Then $a_n^3$ in general looks like

$$\frac{8}{1}, \frac{-1}{1}, \frac{-1}{1}, \frac{8}{2}, \frac{-1}{2}, \frac{-1}{2}, \frac{8}{3}, \frac{-1}{3}, \frac{-1}{3}, \dots$$

which has partial sums $s_{3n} = 6 \sum\limits_{k=1}^n \frac{1}{k}$ diverging.

EDIT: That the converse doesn't hold follows from choosing $a_n = 1/n$. $\sum 1/n^3 < \infty$ but $\sum 1/n = \infty$.

EDIT 2: I'm pretty sure I saw this example somewhere on this website but can't remember where. I'd link to it if I could find it. This was a past qual problem at my university so maybe I learned the answer through other materials.