3

Prove or contradict: if $\sum a_n$ converges then $\sum a_n^3$ converges.

I was able to prove that if $a_n \geq 0$ then the statement is true. But I couldn't prove nor contradict the general case.

catch22
  • 3,057
  • If $\sum a_n$ converges, then $|a_n<1$ for all sufficiently large $n$. And then $a_n^3$ is smaller than $a_n$ and has the same sign. If the sum is absolutely convergent, then you are home. You need to give a little thought to the conditionally convergent case. – almagest Jun 15 '16 at 06:01

1 Answers1

7

False, counterexample: $$a_n = \frac{\epsilon_n}{\sqrt[3]{\lceil n/3 \rceil}} \quad\text{ where }\quad \epsilon_n = \begin{cases}+2,& n \equiv 1, \pmod 3\\ -1, & n \not\equiv 1, \pmod 3\end{cases}$$

It is easy to see $\displaystyle\;\left| \sum_{n=1}^N a_n \right| \le \frac{2}{\sqrt[3]{\lceil N/3 \rceil}} \quad\implies\quad \sum_{n=1}^\infty a_n$ exists and equal to $0$.

However $\displaystyle\;\sum_{n=1}^{3N} a_n^3 = 6\sum_{n=1}^N \frac{1}{n} \approx 6( \log N + \gamma )\;$ diverges logarithmically.

achille hui
  • 122,701