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True or false, prove or find a counterexample. If $\{x_n\}$ is a sequence such that $\{x_n^2\}$ converges, then $\{x_n\}$ converges.

I know how to prove or show $\{x_n\}$ and $\{x_n^2\}$ are convergent using the epsilon/delta definition of convergence. But I don't quite get how to link the two things together. Please help me here. Thank you!

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  • $\begingroup$ Counter example: $x_n = (-1)^n$. $\endgroup$
    – trancelocation
    Commented Feb 28, 2018 at 4:22
  • $\begingroup$ Hints : $(-x)^2 = x^2$, so consider $1,-1,1,-1,...$. What is true ? Well, if $x_n \geq 0$ for all $n$, then indeed the statement is true. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Commented Feb 28, 2018 at 4:22
  • $\begingroup$ I get it now. Thank you! $\endgroup$
    – czhang75
    Commented Feb 28, 2018 at 4:41

2 Answers 2

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The claim is false. Consider $x_n=(-1)^n$, then the sequence starting from $n=0$ alternates 1 and $-1$, so does not converge. But $\{x_n^2\}$ is a constant sequence of ones, so converges.

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If $\{x_n\}$ can change sign, then $(-)^n$ is a counterexample. If not, then the statement is true, since $f(x)=x^{1/2}$ is a continue function.

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