Solution: Let $\varepsilon > 0$ be given. Since $\left( x_n \right) $ is convergent, say converges to $x$, then there exists $n_0\in \mathbb{N} $ such that for every $n\geq n_0$, \[ \left\vert x_n - x \right\vert \lt \varepsilon . \] We claim that $\left( \left\vert x_n \right\vert \right) $ converges to $\vert x \vert $. We recall that \[ \vert \vert a \vert - \vert b \vert \vert \leq \vert a-b \vert . \] Note that for $n\geq n_0$ we have \[ \left\vert \vert x_n \vert - \vert x \vert \right\vert \leq \left\vert x_n - x \right\vert \lt \varepsilon, \] which proves the claim.

The converse need not be true, for example, take $x_n = (-1)^n$. Then $\left\vert x_n \right\vert = 1$, which is convergent but $\left( x_n \right) $ does not.