An example of a series $\sum a_n$ that converges conditionally but $\sum a_n^3$ does not converge

Question

Give an example of a series $\sum a_n$ that converges conditionally but $\sum a_n^3$ does not converge conditionally.

I've come up with an example.

$\frac{1}{\sqrt[3]2}-\frac{1}{2\sqrt[3]2}-\frac{1}{2\sqrt[3]2}+\frac{1}{\sqrt[3]3}-\frac{1}{3\sqrt[3]3}-\frac{1}{3\sqrt[3]3}-\frac{1}{3\sqrt[3]3}+\cdots$.

While the sum of the cubes is

$\frac{1}{2}-\frac{1}{8\cdot 2}-\frac{1}{8\cdot 2}+\frac{1}{3}-\frac{1}{27\cdot 3}-\frac{1}{27\cdot 3}-\frac{1}{27\cdot 3}+\cdots$

Now the series seems to converge to 0, however, I cannot show using an epsilon argument that it does. Also, the sum of the cubes looks like $\frac{1}{4}\cdot \frac{1}{2}+\frac{8}{9}\cdot \frac{1}{3}+ \frac{15}{16}\cdot \frac{1}{4}+\cdots$, so I can see that it diverges, but likewise, cannot supply this with a rigorous argument.

I would greatly appreciate it if anyone can help me with this part.

$\sum a_n^3$ does not means ?? Does not conditionally convergent OR does not convergent ? — Empty, Oct 03 '15 at 19:41
Assuming the former (otherwise the question is misstated), the Harmonic Series is certainly a counterexample. — Travis Willse, Oct 03 '15 at 19:46
How does the Harmonic series work? The series doesn't converge. — nomadicmathematician, Oct 03 '15 at 19:48
You mean to require that $\sum a_n^3$ does not converge at all. @Travis meant the alternating harmonic series, of course. — Ted Shifrin, Oct 03 '15 at 19:49
You would make the example much simpler if you were to use always the $(1,-1/2,-1/2)$ pattern, for all $\sqrt[3]{n}$.. — Lutz Lehmann, Oct 03 '15 at 19:49
@TedShifrin Yes, thank you, that's (of course) what I meant to write. The minus sign was lost when I opted to write out the term in words. — Travis Willse, Oct 03 '15 at 20:48
Possible duplicate of If $\sum a_n$ converges, prove $\sum a_n^3$ converges — , Oct 03 '15 at 21:52
And of Show a convergent series $\sum a_n$, but $\sum a_n^p$ is not convergent — , Oct 03 '15 at 21:54

Empty · Answer 1 · 2015-10-03T19:54:47.170

1

Consider , $a_n=\frac{(-1)^n}{n}$. Then $a_n$ is conditionally convergent. But , $a_n^3=\frac{(-1)^n}{n^3}$ is NOT conditionally convergent ; as it is absolutely convergent.

edited Oct 03 '15 at 19:54

answered Oct 03 '15 at 19:49

Empty

13,012

Well I upvoted. Anyone who downvoted has some 'splainin to do. – zhw. Oct 03 '15 at 19:59
1

Unfair! We are looking at divergent cubes... Absolutely convergent examples from conditionally convergent series are trivial. – Laurent Duval Oct 03 '15 at 21:23
This clearly misses the point of the question – Smiley1000 Apr 03 '23 at 19:13

score 1 · Answer 2 · answered Oct 03 '15 at 19:54

Use the simplified example $$ \frac1{\sqrt[3]2}-\frac1{2\sqrt[3]2}-\frac1{2\sqrt[3]2}+\frac1{\sqrt[3]3}-\frac1{2\sqrt[3]3}-\frac1{2\sqrt[3]3}+\frac1{\sqrt[3]4}-\frac1{2\sqrt[3]4}-\frac1{2\sqrt[3]4}+… $$ Then it is easy to see that this series is conditionally convergent, however the third power series is $3/4$ of the harmonic series in one subsequence of the sequence of partial sums and thus not convergent.

An example of a series $\sum a_n$ that converges conditionally but $\sum a_n^3$ does not converge

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