Prove or disprove that a series is convergent
By James Williams
I was given the following task which I struggle with.
Prove the following statement, or disprove it by giving a counter example:
if $\sum_{n=1}^\infty a_n$ is convergent then $\sum_{n=1}^\infty (a_n)^3$ is also convergent.
I think the statement is true. My intuition is that $|a_n|$<1 starting some index $N$ (since the series converges). Therefore, starting $N$, $|(a_n)^3|$ is smaller than $a_n$, therefore $(a_n)^3$ is "closer" to zero, and since the original series converges, the new one should also converge, yet I can't prove it (nor can I think of a counter example). Any help would be appreciated!
1 Answer
$\begingroup$Consider the sequence:
$$1-\frac{1}{2}-\frac{1}{2}+\frac{1}{\sqrt[3]2}-\frac{1}{2\sqrt[3]2}-\frac{1}{2\sqrt[3]2}+\frac{1}{\sqrt[3]3}-\frac{1}{2\sqrt[3]3}-\frac{1}{2\sqrt[3]3}+\frac{1}{\sqrt[3]4}-\frac{1}{2\sqrt[3]4}-\frac{1}{2\sqrt[3]4}+\ldots$$
It clearly converges to $0$, but the sequence of its cubes does not (by combining terms you can see that the sequence of its cubes becomes $3/4$ times the harmonic series).
$\endgroup$ 3
