Comparative status of $\sum a_n$ and $\sum {a_n}^{[2]}$.

Question

Let $(a_n)_n$ be a sequence of real terms.

I'm interested in finding a counter-example to the assertion $$\sum\limits_{n=0}^{\infty}a_n\text{ converges }\implies\sum\limits_{n=0}^{\infty}{a_n}^{[2]}\text{ converges}$$

Where $x^{[2]}=x^2\times \operatorname{sgn}(x)=x\,|x|$ is the odd square function.

• If the sequence has only positive terms $a_n\ge 0$ or if $\sum a_n$ is absolutely convergent then it can be proven that $\sum {a_n}^2$ is absolutely convergent as well.

$\sum_\limits{n=0}^{\infty} a_n$ converges $\implies \sum_\limits{n=0}^{\infty} a_n^2$ converges

Prove that if $\sum{a_n}$ converges absolutely, then $\sum{a_n^2}$ converges absolutely

• If the CAS (criteria of alternated series) is verified for $\sum a_n$ then it is also verified for $\sum {a_n}^{[2]}$ so both series are convergent.

Indeed if $|a_{n+1}|<|a_n|$ then since $x\mapsto x^2$ is increasing on $[0,+\infty[$ then $|a_{n+1}|^2\le|a_n|^2$

Also since $\operatorname{sgn}(a_n)=\operatorname{sgn}({a_n}^{[2]})$ they are both alternated series.

• Finally since $x\mapsto x^{[2]}$ is not linear in a neighbourhood of zero then it is not a convergence preserving function according to this result:

$f$ such that $\sum a_n$ converges $\implies \sum f(a_n)$ converges

The set of functions which map convergent series to convergent series

So it should be possible to exhibit a sequence $(a_n)_n$ such that $\sum a_n$ is convergent while $\sum {a_n}^{[2]}$ is not.

Unfortunately I haven't been very successful at expliciting such a sequence, in fact I was wondering if the theorem was not true after all until I found the postulate above regarding CP functions.

Do you know any counter-examples for the assertion at the beginning of the post ?

Answer 1

Nice research. I can tell you've been searching. Here is a counterexample for the general $a_n$ converges implies $a_n^{[k]}$ converges (for $k \neq 1$). We can construct the series from groups of threes as follows. $$\frac{1}{n^{1/k}} + \frac{1}{n^{1/k}} - \frac{2}{n^{1/k}}$$ This series converges as the sum of each set of three is $0$ and the limit of the terms is also $0$ as $n\to\infty$. But when we take the $k$th power (keeping the sign) of each term, we end up with a constant multiple of the harmonic series. $$\frac{1}{n} + \frac{1}{n} - \frac{2^k}{n}=(2-2^k)\frac{1}{n}$$ And as we all know, the harmonic series diverges.