# Is there a non alternating sequence that diverges but converges when squared?

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

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?

• $a_n$ 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) – Yanko Oct 22 '18 at 13:04
• Do you mean a positive sequence ? – Yves Daoust Oct 22 '18 at 13:10
• 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. – user605734 MBS Oct 22 '18 at 13:16
• 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? – 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$$\lim_{n\to\infty}{a_n}^2=l\implies\lim_{n\to\infty} a_n=-\sqrt l.$$
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 $$-1$$.
• @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. – Andrea Mori Oct 22 '18 at 13:50
• then: $a_n = 1 : n\neq k! , a_n = -1 : n = k! \forall n \in N$ – user605734 MBS Oct 22 '18 at 13:54