TITLE: $\sum^{\infty}_{n=1}x_n$ converges $\Rightarrow$ $\sum^{\infty}_{n=1}x_n^3$ converges also?
QUESTION [7 upvotes]: Let the series $\sum^{\infty}_{n=1}x_n$ converge. $\{x_n\}_{n=1}^{\infty} \in \mathbb R$
Does $\sum^{\infty}_{n=1}x_n^3$ converge too?

I tried to find some counter-examples but found none. I tried to prove this also, but I can't... I don't know even if it's true or not.

REPLY [4 votes]: No, if the signs of the $x_n$ are not the same; here's a fairly typical counterexample.  Define the series $\sum_nx_n$ as follows. Its positive terms are $1/n^{1/3}$ for all natural numbers $n$, each of these terms occurring exactly once. Its negative terms are $-1/n^{4/3}$, each occurring exactly $n$ times, immediately after the corresponding positive term $1/n^{1/3}$.  In other words, the series consists of blocks of consecutive terms, where the $n$th black consists of $1/n^{1/3}$ followed by $n$ consecutive occurrences of $-1/n^{4/3}$. Notice that each block has sum $0$, from which it easily follows that the whole series converges to $0$.  Now consider what happens when you cube each term. The $n$th block of the new series consists of a positive term $1/n$ followed by $n$ consecutive negative terms $-1/n^4$.  The $n$ negative terms in the $n$th block add up to $-1/n^3$ which is far from cancelling the positive term $1/n$.  Indeed, except for the first block or two, the sum of the $n$th block will be positive and larger than $1/(2n)$.  Since the harmonic series diverges, this means that the series of cubes diverges to $+\infty$.