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

Question

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!

Answer 1

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.

Answer 2

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.