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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!

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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).

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  • $\begingroup$ Since there are two identical answers to this question. I wonder if your solution is from yourself or from a solution book you happened to have? $\endgroup$ – DeepSea Aug 3 '15 at 3:55
  • $\begingroup$ It's my own. The motivation is that when you try to prove the statement using your method, you realize that you run into a problem with positive and negative numbers, so you try to make all the negative numbers "closer" to 0 than the positive numbers, so that cubing them makes their absolute value even smaller. $\endgroup$ – msinghal Aug 3 '15 at 4:44
  • $\begingroup$ Good job. +1 for innovative answer. $\endgroup$ – DeepSea Aug 3 '15 at 4:45

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