$\{a_n\}_{n\in N}$ diverges, can the sequence $\{a_n^2\}_{n\in N}$ converge?

Answer: yes

For example:

$\{(-1)^n\}_{n\in N}$ Does not have a real limit, therefore diverges, and $\{(-1)^{2n}\}_{n\in N}$ equals to the constant series $\{1\}_{n\in N}$, which obviously converges.

But $\{(-1)^n\}_{n\in N}$ is also an alternating sequence, Is there a non alternating sequence that diverges but converges when squared?

  • 1
    $\begingroup$ $a_n$ must get infinitely many positive values and infinitely many negative values in order to be a counter-example. You should define the notion "alternating sequence" because for example $1,1,-1,1,1,-1,1,1,-1,...$ is a counter-example yet it is not "alternating" in the usual sense (i.e. positive follows negative) $\endgroup$ – Yanko Oct 22 '18 at 13:04
  • $\begingroup$ Do you mean a positive sequence ? $\endgroup$ – Yves Daoust Oct 22 '18 at 13:10
  • $\begingroup$ Not necesarily, I would like one that does not alternate, and the previous example is actually alternating (with a period 3) I mean, I can take the sequence $a_n = 2 + (-1)^n$ , every item is possitive but it still alternates. $\endgroup$ – user605734 MBS Oct 22 '18 at 13:16
  • $\begingroup$ The sequence {-1,1,1,−1,1,1,−1,1,1,−1,...} can be expressed as $a_n = (-1)^{n-\lfloor \frac{n}{3} \rfloor}$, right? $\endgroup$ – user605734 MBS Oct 22 '18 at 13:24

If a sequence $(a_n)_{n\in\mathbb N}$ is such that $a_n\geqslant0$ if $n\gg1$ and if furthermore it diverges, then $({a_n}^2)_{n\in\mathbb N}$ diverges too. This is so because$$\lim_{n\to\infty}{a_n}^2=l\implies\lim_{n\to\infty} a_n=\sqrt l.$$And if $a_n\leqslant0$ if $n\gg1$, then$$\lim_{n\to\infty}{a_n}^2=l\implies\lim_{n\to\infty} a_n=-\sqrt l.$$


There are many (uncountably many?) ways to obtain a non-alternating sequence and in fact a non-periodic sequence as you wish.

Possibly the simplest is to choose any irrational number $\alpha\in (0,1)$ and consider its binary expansion, say something like $$ \alpha=0,00110111010011110010000000101111... . $$ Then consider the sequence of digits when wherever you read $0$ put $-1$.

  • $\begingroup$ How can I express this sequence mathematically? I understand how you get it BTW thanks $\endgroup$ – user605734 MBS Oct 22 '18 at 13:38
  • $\begingroup$ @user605734MBS: you can express it simply by saying $\alpha$. The number fixes its expansion in base $2$. If we wish something apparently more explicit you may consider the sequence $a_n$ where $a_n=1$ except when $n=k!$ where you put $a_n=-1$ (so for $n=1$, $2$, $6$, $24$, $120$ and so on). By the way, if this came from some $\alpha$ following the recipe above then you know that $\alpha$ is transcendental. $\endgroup$ – Andrea Mori Oct 22 '18 at 13:50
  • $\begingroup$ then: $a_n = 1 : n\neq k! , a_n = -1 : n = k! \forall n \in N$ $\endgroup$ – user605734 MBS Oct 22 '18 at 13:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.