Example of a bounded sequence $(x_n)$ where $(\lim\sup x_n)^2\neq \lim \sup (x_n^2)$

real-analysis sequences-and-series limsup-and-liminf

My question is exactly as the title states. I'm struggling to come up with an example of a bounded sequence $(x_n)$ where $(\lim\sup x_n)^2\neq \lim \sup (x_n^2)$. Anyone have any simple examples? Thanks in advance.

Does this work? $(x_n)=(-1,-2,-1,-2,...)$. Then $(\lim\sup x_n)^2=(-1)^2=1$ but $(x_n)^2=(1,4,1,4,...)$ so $\lim \sup (x_n^2)=4$


Does this work? $(x_n)=(-1,-2,-1,-2,...)$. Then $(\lim\sup x_n)^2=(-1)^2=1$ but $(x_n)^2=(1,4,1,4,...)$ so $\lim \sup (x_n^2)=4$

Yes, that works perfectly well, and is essentially what I was hinting at with my comment. As an extension, try to show $$\limsup (x_n^2) = \max\left((\limsup(x_n))^2,(\liminf(x_n))^2\right)$$ (You can also try to find a formula for $\liminf(x_n^2)$ but it's a little more difficult to write out)

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