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$\{a_n\}_{n\in N}$ 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?

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    $\begingroup$ $a_n$ 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
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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$$\lim_{n\to\infty}{a_n}^2=l\implies\lim_{n\to\infty} a_n=-\sqrt l.$$

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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 $-1$.

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  • $\begingroup$ 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

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