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Per the title, if $\sum_{1}^{\infty}(a_n)^3$ diverges, does this imply that $\sum_{1}^{\infty}(a_n)$ diverges?

I'd appreciate hints (!) for dealing with this excercise.

EDIT Per the contrapositive, it is not given that $a_n$ converges absolutely, or that it is nonnegative for all $n$.

Thank you!

ro44
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  • Are you assuming the $a_n$ are all non-negative real? – Geoff Robinson Jan 05 '12 at 15:15
  • @Geoff: I am not assuming that. – ro44 Jan 05 '12 at 15:19
  • For the sums to converge, you need $|a_n| \to 0$. Thus at some point, you'll have $0 < |a_n| < 1$. Now compare $a_n$ and $a_n^3$. – Dario Jan 05 '12 at 15:20
  • @Dario: I can compare $|a_n|$ and $|a_n^3|$, but as it's not given that the $|a_n|$ converges, does this help me? (Assuming you're speaking about the contrapositive) – ro44 Jan 05 '12 at 15:26
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    Yes, prove the contrapositive. I.e. show that if $\sum a_n$ converges, $\sum a_n^3$ will converge. – Dario Jan 05 '12 at 15:28
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    @Dario: I can see why that statement is true when $a_n$ converges absolutely, however, not when all we know is that it converges. Can you explain your hint further? – ro44 Jan 05 '12 at 15:32
  • @GeoffRobinson: Clarification, I'm assuming they're all real but not that they're nonnegative! – ro44 Jan 05 '12 at 15:54
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    It's not true if the $a_n$ are allowed to be complex. You can take $a_{3k+j} = \frac{1}{\sqrt[3]k} \zeta_3^j$, where $\zeta_3$ is a primitive cubed root of $1$ and $j=0,1,2$. Then $a_{3k}+a_{3k+1}+a_{3k+2}=0$, but $a_{3k}^3+a_{3k+1}^3+a_{3k+2}^3 = \frac{3}{k}$ – Thomas Andrews Jan 05 '12 at 15:54
  • @ThomasAndrews I just stumbled here, more than 4 years later, so maybe you won't see this. Your counterexample does not seem to work. Sure, the sum of three consecutive terms is zero, but that does not mean that the partial sums are all zero. Indeed, the partial sums $s_n$ for the series $\sum a_n$ do not converge either, since it is an oscillating sequence. – bartgol Apr 15 '16 at 19:01
  • @bartgol The partial sums are not zero, but they are less, in absolute value, than $$\frac{2}{\sqrt[3]{\lfloor n/3\rfloor}}$$ If I hadn't included the $\frac{1}{\sqrt[3]{k}}$ in the formula, it wouldn't converge to zero, but as defined, it does. – Thomas Andrews Apr 15 '16 at 19:36
  • Uh, right. Thanks. – bartgol Apr 17 '16 at 22:49

4 Answers4

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In The American Mathematical Monthly, Vol. 53, No. 5, (May, 1946), pp. 283-284, you will find N. Fine's solution to G. Polya advanced problem 4142:

Let $C$ be an arbitrary subset of the positive integers ($C$ may be finite or infinite). Then there is a sequence $a_1,a_2,a_3,\dots$ of real numbers (of course, depending on $C$) such that for any positive integer $l$, $$ \sum {a_n}^{2l-1} $$ converges if and only if $l\in C$.

In particular, we can choose $C$ so that the corresponding sequence gives us an example where $\sum a_n^3$ diverges and $\sum a_n$ converges.

Two quick observations: First, in general, the $a_n$ are not going to be nonnegative. This is because if $\sum b_n$ converges then $b_n\to 0$, so if $\sum {a_n}^{2k-1}$ converges and all the $a_n$ are nonnegative, we have some $N$ such that for $n\ge N$ we have that $0\le a_n<1$. But then ${a_n}^{2l-1}={a_n}^{2k-1}{a_n}^{2(l-k)}\le {a_n}^{2k-1}$ for all $n\ge N$ and all $l\ge k$. In particular, if $\sum a_n$ converges, then so does $\sum {a_n}^3$.

Second, note that this only deals with odd exponents. This cannot be helped: For the same reason as in the previous paragraph, if $\sum {a_n}^{2k}$ converges, then so does $\sum {a_n}^{2l}$ for any $l\ge k$.

Fine's solution is constructive, by the way. Following his method, you produce an explicit example of a sequence $(a_n)$. In particular, his method gives that if $$ a_1=1, a_2=a_3=-\frac12, a_4=\frac1{\root3\of2},a_5=a_6=-\frac12\frac1{\root3\of2},a_7=\frac1{\root3\of3},\dots $$ (where for each $n$ you list $\displaystyle\frac1{\root3\of n}$ immediately followed by two $\displaystyle-\frac12\frac1{\root3\of n})$ then $\sum a_n$ converges (to 0) while $\sum {a_n}^3$ diverges (essentially because the harmonic series diverges).

In fact, in this example we have that all of $\sum {a_n}^5,\sum {a_n}^7,\dots$ converge as well.

More general patterns can be obtained if we allow the $a_n$ to be complex numbers. Variants of this problem have appeared in the Monthly a few times over the years. It would be nicer if this result were better known.

(Since this is tagged as homework, I'll leave out the details of Fine's solution.)

Andrés E. Caicedo
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  • Probably should have started with the explicit counter-example for this problem, since it is fairly easy to understand, although the general theorem is really nice. – Thomas Andrews Jan 05 '12 at 16:01
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Consider a series like this: The terms are in groups of $3$. The $n$th group has two positive then one negative term: $$ \frac{1}{n^{1/3}} + \frac{1}{n^{1/3}} - \frac{2}{n^{1/3}} $$ This series then converges, but only conditionally. However, after you cube the terms, the $n$th group is: $$ \frac{1}{n} + \frac{1}{n} - \frac{8}{n} $$ So the series of cubes diverges to $-\infty$.

GEdgar
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$$\left[\left(1-{1\over2}-{1\over2}\right)+\cdots+\left(1-{1\over2}-{1\over2}\right)\right]+\left[\left({1\over2}-{1\over4}-{1\over4}\right)+\cdots+\left({1\over2}-{1\over4}-{1\over4}\right)\right]+\cdots$$

will clearly converge to $0$, while the sum of the cubes, $1-{1\over8}-{1\over8}+\cdots$, can be made to diverge if there are more and more terms for the "$\cdots$" within each set of square brackets.

Barry Cipra
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hint: assume $a_n\geq0$ and $\sum a_n$ converges. then $a_n\to0$ and $0\leq a_n^3\leq a_n$ for large enough $n$ (the cube of a number between zero and one is smaller than the number itself). what does this say about $\sum a_n^3$?

yoyo
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  • By the comparison test, it says that $a_n^3$ converges. But does this generalise for when we don't assume that $a_n\geq 0$? – ro44 Jan 05 '12 at 15:31