General infinite series convergence or divergence. 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.
 A: 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 $a_n = \frac{\cos \frac{2\pi n}3}{\sqrt[3] n}$. 
A: 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.
A: 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*}
