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This is an exercise that was given to me, and I would like to emphasize that I am not looking for an answer, but rather a pointer or a hint in the right direction, since it appears I've reached a bit of a dead end. We are looking at sequences over the real numbers (at least that's my assumption, and that's where I've been looking).

I've come with various results to help me try such a sequence:

  • The ratio test has to be inconclusive, as does the root test.
  • Since $a_n$ converges to $0$, so does $a_n^3$
  • The sequences $|a_n|$ cannot be monotone decreasing, otherwise $\sum a_n^3$ will converge.

Other various things I've noticed:

  • It's very easy to find $a_n$ such that $a_n$ converges but $a_n^2$ diverges. Is it possible to use that fact? Let $u_n$ be such a sequence. I tried to construct a sum of two sequences so that when we take the third power, $u_n^2$ appears in the binomial expansion. That "naive" approach didn't get me too far however.
  • We can make the following observation: \begin{align} \sum a_n^3 &= a_1^3 + a_2^3+a_3^3+a_4^3\cdots \\ & = (a_1+a_2)((a_1-a_2)^2 + a_1a_2) + (a_3+a_4)((a_3-a_4)^2 + a_3a_4) +\cdots \end{align} This hasn't taken me anywhere.

Other things I've thought about:

  • I think a natural form for the sequence would be the product of two functions, $f$ and $g$, such that one converges to $0$ much faster than the other. By using the comparison test, we could try to require that $(fg)^3 > k_n$ for all n for some sequence $k_n$ whose associated series is divergent. I think it is unlikely we could find such a product though, since if $fg$ plummets fast enough, then $(fg)^3$ will plummet even faster.

    • I've tried a few things for lack of better options, using exp, trig functions, polynomials, none of my tries have worked so far, almost always as a product of two functions.

If someone could suggest a possible step forward, I would be very grateful!

Quantaliinuxite
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    Hint:$$a_n=\frac{(-1)^n}{\sqrt n}$$This example has $\sum a_n$ converging and $\sum a_n^2$ diverging (and in particular, $a_n^2$ is always positive). What happens if you took $$a_n=\frac{b_n}{\sqrt[3]n}$$so that $\sum a_n$ converges, $|b_n|=1$ and $a_n^3$ were always positive? – Simply Beautiful Art Oct 17 '17 at 01:15
  • @SimplyBeautifulArt It seems that there is no choice but to choose $b_n = 1$? –  Oct 17 '17 at 01:19
  • @SimplyBeautifulArt Cubing is not like squaring. Negative things stay negative. – 2'5 9'2 Oct 17 '17 at 01:19
  • @SimplyBeautifulArt Been there done that! – Quantaliinuxite Oct 17 '17 at 01:19
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    Jeez, why is everyone pinging me!? – Simply Beautiful Art Oct 17 '17 at 01:20
  • @JohnMa No. You need to think more... *complex* – Simply Beautiful Art Oct 17 '17 at 01:20
  • @alex.jordan See above – Simply Beautiful Art Oct 17 '17 at 01:20
  • @SimplyBeautifulArt I am pretty sure I'm supposed to look at sequences over the reals – Quantaliinuxite Oct 17 '17 at 01:21
  • @SimplyBeautifulArt See the question again..... –  Oct 17 '17 at 01:21
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    If you want more than a hint, see here: https://math.stackexchange.com/questions/2216555/general-infinite-series-convergence-or-divergence – carmichael561 Oct 17 '17 at 01:25
  • @carmichael561 Oh right, silly me. I forgot, just take real and imaginary parts of what my hint would provide. Doh! – Simply Beautiful Art Oct 17 '17 at 01:26
  • $\sum_{n=2}^\infty \frac{(-1)^n}{ \log n}$ converges. Replace each $\frac{1}{\log 2n}$ term by $2n$ terms $\frac{1}{2n\log 2n}$. The obtained series $\sum_{k=2}^\infty a_k$ converges and $\sum_{k=2}^\infty a_k^3$ diverges. – reuns Oct 17 '17 at 01:31
  • @SimplyBeautifulArt There is a simpler way to proceed here than that suggested in the link. I don't think it should be marked as a duplicate for that reason. – zhw. Oct 17 '17 at 04:58
  • Consider the series

    $$1-\frac{1}{2}\cdot 1-\frac{1}{2}\cdot 1 + \frac{1}{2^{1/3}}- \frac{1}{2}\cdot \frac{1}{2^{1/3}} - \frac{1}{2}\cdot \frac{1}{2^{1/3}} + \frac{1}{3^{1/3}}- \frac{1}{2}\cdot \frac{1}{3^{1/3}} - \frac{1}{2}\cdot \frac{1}{3^{1/3}} + \cdots$$

    which converges to $0.$ Cubing the series gives ...

     – zhw. Oct 17 '17 at 05:11
  • $$1- \frac{1}{8}-\frac{1}{8}+ \frac{1}{2}- \frac{1}{8}\cdot \frac{1}{2}- \frac{1}{8}\cdot \frac{1}{2} + \frac{1}{3}- \frac{1}{8}\cdot \frac{1}{3}- \frac{1}{8}\cdot \frac{1}{3} + \cdots $$ $$= \frac{3}{4}\cdot 1 +\frac{3}{4}\cdot \frac{1}{2} + \frac{3}{4}\cdot \frac{1}{3} + \cdots = \infty$$ – zhw. Oct 17 '17 at 05:14
  • @zhw. If you think there are better answers than those found in a dupe, you should add your answers over there. – Simply Beautiful Art Oct 17 '17 at 10:47

2 Answers2

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You are looking for hints only, so here is one. Since the sum of $a_n$ converges, what does this tell you about the relationship between $|a_n|$ and $|a_n^3|$ once $n$ is large enough?

2'5 9'2
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Can you think of a function $f$ such that $\int f dx$ converges and $\int f^3 dx$ diverges (with the integral over some set, say $\mathbb{R}$ or $[0,1]$)?