Let $(a_k)$ a sequence of real number. How can I show that $$\sum_{k=1}^\infty a_k<\infty \implies \sum_{k=1}^\infty a_k^3<\infty.$$

If $(a_k)$ has constant sign, it's obvious. But unfortunatly it's not supposed, so comparaison criterion doesn't work.

Then I tried by contradiction : if it doesn't converge, then there are subsequence $(n_k)$ and $(m_k)$ s.t. $(S_{n_k})$ and $(S_{m_k})$ has not the same limit (where $S_n=\sum_{k=0}^n a_k^3$. But I can't conclude in that way.

Any idea ?

  • $\begingroup$ I do not understand your notation. Are you saying that the original series converges, and want to conclude that the second series converges as well? $\endgroup$ – Andrés E. Caicedo Dec 23 '17 at 14:25
  • $\begingroup$ @AndrésE.Caicedo : I think that it's a common notation to says that $\sum_{n}a_n$ converge implies $\sum_{n}a_n^3$ also converge... $\endgroup$ – Surb Dec 23 '17 at 14:27
  • $\begingroup$ Anyway, if my reading is correct, then what you are trying to prove is false. Of course, it is true is the $a_n $ are nonnegative. $\endgroup$ – Andrés E. Caicedo Dec 23 '17 at 14:28
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    $\begingroup$ I posted a solution. By the way, it is ambiguous to write $\sum a_n<+\infty$ when what you really mean is that the series converges. For instance, this notation allows for the possibility that $\sum a_n$ diverges to $-\infty$. $\endgroup$ – Andrés E. Caicedo Dec 23 '17 at 14:53
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    $\begingroup$ If the $a_k$ are not positive then it's really not so clear what "$\sum a_k<\infty$" means - if you meant to say that $\sum a_k$ converges much better notation would be "$\sum a_k$ converges". $\endgroup$ – David C. Ullrich Dec 23 '17 at 14:55

For a sequence $(a_n)$ of real numbers, it is not true that the convergence of $\sum a_n$ implies the convergence of $\sum a_n^3$. Of course this holds if the $a_n$ are nonnegative, by a simple comparison argument: For $m$ large (say, $m>N$), $a_m<1$ since in fact the convergence of $\sum a_n$ implies that $a_n\to 0$. But then $0\le a_m^3\le a_m$ so $\sum_{m>N}a_m^3\le \sum_{m>N}a_m$ and the result follows.

Similarly, if $\sum a_n^k$ converges for some even number $k$, then $\sum a_n^l$ converges as well for any $l>k$, since in fact $\sum |a_n|^l$ converges by the argument from the previous paragraph.

So, the issue in the problem at hand is that we are dealing with odd powers. In 1944, George Pólya posted the following question to the American Mathematical Monthly:

Let $C$ be an arbitrarily given class of positive integers ($C$ could be infinite). Show that there is a sequence $(a_n)$ of real numbers such that for any positive integer $l$, $\sum a_n^{2l-1}$ converges if and only if $l\in C$.

In particular, there are sequences such that $\sum a_n$ converges and $\sum a_n^3$ diverges. For the general solution to his problem, I refer you to N. J. Fine's solution in Vol. 53, No. 5, (May, 1946), pp. 283-284 of the Monthly.

Here, let me just give an example. Consider the sequence $a_1,a_2,a_3,\dots=$ $$1^{-1/3}, -\frac12\cdot 1^{-1/3},-\frac12\cdot 1^{-1/3},2^{-1/3},-\frac12\cdot 2^{-1/3},-\frac12\cdot 2^{-1/3},\dots,n^{-1/3},-\frac12\cdot n^{-1/3},-\frac12\cdot n^{-1/3},\dots$$ It should be readily apparent that $\sum a_n$ converges (to 0) and that $\sum a_n^3$ diverges (since the harmonic series diverges). In fact, you may want to verify that $\sum a_n^{2k-1}$ converges for all $k\ne 2$. Fine's solution to Pólya's problem proceeds by finding a sequence that behaves that way for any odd positive integer $m$: A sequence such that $\sum a_n^{2k-1}$ converges if and only if $2k-1\ne m$. Once we have these building blocks, it is not hard to see how to combine them to ensure convergence or divergence at any odd number as desired.

  • $\begingroup$ Possible generalization? For any smooth, increasing bijection $f:\mathbb{R} \to \mathbb{R}$ with $f(0) = 0$ and $f = O(x^3)$ as $x \to 0$, there is a sequence $\{a_n\}$ such that $\sum a_n$ converges but $\sum f(a_n)$ diverges. $\endgroup$ – MathematicsStudent1122 Dec 23 '17 at 15:02

Counter-example: You may concatenate lumps of terms of the following form:

For $k\geq 1$ denote $m_k=\lfloor k^{1/6}\rfloor$ and define $N_1=1$, $N_{k+1}= N_k + m_k+1$. For $0\leq j< m_k$ set $$ x_{N_k+j} = -\frac {1}{k^{1/3}} , \; 0\leq j< m_k, \;\;\; x_{N_k+m_k} = \frac{m_k}{k^{1/3}} \sim k^{-1/2}$$ Then the sum $$ \sum_{j=0}^{m_k} x_{N_k+j} = 0 $$ But $$ \sum_{j=0}^{m_k} x^3_{N_k+j} \sim \frac{1}{k^{5/6}} $$ Summing over $k$ the latter goes to infinity. It follows that $\sum_{n\geq 1} x_n$ converges (conditionally) whereas $\sum_n x_n^3$ does not.


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