# If $\sum_{1}^{\infty}(a_n)^3$ diverges, does $\sum_{1}^{\infty}(a_n)$?

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!

#### Solutions Collecting From Web of "If $\sum_{1}^{\infty}(a_n)^3$ diverges, does $\sum_{1}^{\infty}(a_n)$?"

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.)

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 $-\infty$.

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 $\sum a_n^3$?

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

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.