Suppose the infinite series $\sum a_n$ converges ($a_n \in \mathbb{R}$ not all positive or negative).
(1) Is it true that $\sum a_n^2$ converges?
(2) Is it true that $\sum a_n^3$ converges?
For (1), the answer is no. I found the counterexample $a_n = (-1)^n / \sqrt{n}$ which converges by alternating series, but $\sum a_n^2 = \sum1/n $ is divergent.
I'm unsure about proving truth of (2) and can't seem to find a counterexample.
Thanks for help.
For (2): With slight modification of @carmichael561's answer, we use the triple angle formula for cosine function as follows: $$ \cos 3\theta= 4\cos^3 \theta - 3\cos \theta. $$ Let $\theta = \frac{2\pi n}3 $. Then $$ 1=4\cos^3{\frac{2\pi n}3}-3\cos\frac{2\pi n}3. $$ By Dirichlet test, $$ \sum_{n=1}^{\infty} {\frac{\cos \frac{2\pi n}3}{\sqrt[3] n}} \ \ \textrm{and} \ \ \sum_{n=1}^{\infty} {\frac{\cos \frac{2\pi n}3}{n}} \ \ \textrm{converges,} $$ but since $\cos^3{\frac{2\pi n}3}=\frac{1+3\cos\frac{2\pi n}3}4$, $$ \sum_{n=1}^{\infty} \frac{\cos^3 \frac{2\pi n}{3}}{n} \ \ \textrm{diverges.} $$ Therefore, this provides a counterexample for (2) with $a_n = \frac{\cos \frac{2\pi n}3}{\sqrt[3] n}$.
Claim (2) is also false, as you can show using an analogue to your counterexample for claim (1):
Let $\omega=e^{\frac{2\pi i}{3}}$, so that $\omega^3=1$, and let $$a_n=\frac{\omega^n}{\sqrt[3]{n}}$$
Then $\sum_na_n$ converges by Dirichlet's test for convergence, because $1+\omega+\omega^2=0$ so the partial sums $\sum_{n=1}^N\omega^n$ are bounded. However, $a_n^3=\frac{1}{n}$, so $\sum_na_n^3$ diverges.
Here is another example for $(2)$: \begin{align*} \sum_{n=0}^{\infty} a_n = 1 + \frac{1}{2 2^{\frac{1}{3}} } + \frac{1} {2 2^{\frac{1}{3}} } - \frac{1}{2^{\frac{1}{3} }} + \dots + \underbrace{\frac{1}{n n^{\frac{1}{3}} } + \dots + \frac{1} {n n^{\frac{1}{3}} }}_{n} - \frac{1}{n^{\frac{1}{3} }} + \dots \end{align*}
