T/F: If $\sum_{n=1}^{\infty}a_{n}$ converges, then $\sum_{n=1}^{\infty}(a_{n})^{3}$ converges. My math GSI proposed this question to the class today, and even he doesn't know the answer!
(This should be answered separately for absolute and conditional convergence)
 A: Not necessarily. Consider the series
$$\frac{1}{\sqrt[3]{2}}-\frac{1}{2\sqrt[3]{2}}-\frac{1}{2\sqrt[3]{2}}+\frac{1}{\sqrt[3]{3}}-\frac{1}{2\sqrt[3]{3}}-\frac{1}{2\sqrt[3]{3}}+\frac{1}{\sqrt[3]{4}}-\frac{1}{2\sqrt[3]{4}}-\frac{1}{2\sqrt[3]{4}}+\frac{1}{\sqrt[3]{5}}-\frac{1}{2\sqrt[3]{5}}-\frac{1}{2\sqrt[3]{5}}+\cdots.$$
It is clear that this converges. However, cubing basically gives us three-quarters of the harmonic series. 
Remark: If the original series converges absolutely, then the sum of the cubes is trivially absolutely convergent. The $|a_n|$ are bounded, say by $b$. then $|a_n^3|\lt |a_n|b^2$, so by Comparison $\sum |a_n^3|$ converges. 
A: Here is a general construction which may prove useful. Fix some $k\geqslant2$, some real numbers $(c_i)_{1\leqslant i\leqslant k}$ and some real number $a\gt0$, and, for every $n\geqslant1$, define
$$
x_{n}=\frac{c_i}{n^a}\quad\text{if}\quad n=i\pmod{k}.
$$
Then:

If $c_1+c_2+\cdots+c_{k}=0$, the series $\sum\limits_nx_n$ converges, for every $a\gt0$. If $c_1+c_2+\cdots+c_{k}\ne0$, the series $\sum\limits_nx_n$ converges if and only if $a\gt1$.

To check this, introduce the $j$th slab $s_j$ as
$$
s_j=\sum\limits_{n=jk+1}^{jk+k}x_n.
$$
Then,
$$
s_j=\sum_{i=1}^{k}\frac{c_i}{(jk)^a}\left(1+\frac{i}{jk}\right)^{-a}=\sum_{i=1}^{k}\frac{c_i}{(jk)^a}\left(1+O\left(\frac1{j}\right)\right).
$$
If $c_1+c_2+\cdots+c_{k}=0$ the first terms disappear and one is left with
$$
s_j=O\left(\frac1{j^{1+a}}\right),
$$
hence the series
$$
\sum_js_j
$$
converges. Since $x_n\to0$ and the slabs are composed of a uniformly bounded number of terms, this proves that the series $\sum\limits_nx_n$ converges. If $c_1+c_2+\cdots+c_{k}\ne0$,
$$
s_j\sim\frac{1}{(jk)^a}\sum_{i=1}^{k}c_i=\Theta\left(\frac1{j^a}\right),
$$
hence the series $\sum\limits_ns_n$ converges if and only if $a\gt1$ and the result follows.
Noting that, for every positive integer $\ell$, 
$$
(x_n)^\ell=\frac{(c_i)^\ell}{n^{a\ell}}\quad\text{if}\quad n=i\pmod{k},
$$
one can apply both parts of the result above to deduce:

If $c_1+c_2+\cdots+c_{k}=0$ but $(c_1)^\ell+(c_2)^\ell+\cdots+(c_{k})^\ell\ne0$ and if $a\ell\leqslant1$, then the series $\sum\limits_nx_n$ converges but the series $\sum\limits_n(x_n)^\ell$ diverges.

Example: choose $\ell=3$, $0\lt a\leqslant\frac13$, $k=3$, and $(c_1,c_2,c_3)=(3,-1,-2)$. Then $(c_1)^3+(c_2)^3+(c_3)^3=18\ne0$ hence $\sum\limits_nx_n$ converges but $\sum\limits_n(x_n)^3$ diverges. 
The same example but with $0\lt a\leqslant\frac1{2p+1}$ yields a series such that $\sum\limits_nx_n$ converges but $\sum\limits_n(x_n)^{2q+1}$ diverges for every $1\leqslant q\leqslant p$.
A: let $$a_{n}=\dfrac{\cos{\dfrac{2n\pi}{3}}}{\sqrt[3]{n}}$$
By Dirichlet test,we easy to find this series
$$\sum_{n=1}^{\infty}\dfrac{\cos{\dfrac{2n\pi}{3}}}{\sqrt[3]{n}}$$ is convergent:
But other hand 
Use $$\cos^3{x}=\dfrac{1}{4}(\cos{(3x)}+3\cos{x})$$
so
$$a^3_{n}=\dfrac{1}{4}\left(\dfrac{1}{n}+\dfrac{3\cos{\dfrac{2n\pi}{3}}}{n}\right)$$
so
$$\sum_{n=1}^{\infty}a^3_{n}$$ is 
disconvergent:
