For any real numbers $a,b$ we have $\vert \vert a \vert - \vert b \vert \vert \leq \vert a - b \vert $
Solution: Let $\epsilon > 0$ be given. Since the sequence $\{ s_n \} $ is convergent, let the limit be $s$. So we will find $n_0 \in \mathbb{N} $ such that for $n \geq n_0$,
\begin{equation}\label{eq:31Mar2025-1}
\left\vert s_n - s \right\vert < \epsilon .
\end{equation}
Now for $n\geq n_0$ consider,
\begin{align*}
\left\vert \vert s_n \vert - \vert s \vert \right\vert & \leq \left\vert s_n - s \right\vert < \epsilon .
\end{align*}
Thus, $\{ \vert s_n \vert \} $ converges to $s$.
The converse need not be true. For example, take $s_n = (-1)^n$, then
\[
\vert s_n \vert = 1 \rightarrow 1,
\]
but $\{ s_n \} $ is not convergent. Thus, the converse need not be true.
Let us prove the inequality that we used here, that is, for any real numbers $a,b$ we have $\vert \vert a \vert - \vert b \vert \vert \leq \vert a - b \vert $. Note that for any real numbers $a,b$, we have
\begin{align*}
& \vert a \vert - \vert (-b) \vert \leq \vert a + (-b) \vert \quad \text{ and } \quad \vert b \vert - \vert (-a) \vert \leq \vert b + (-a) \vert \\
\implies & \vert a \vert - \vert b \vert \leq \vert a - b \vert \quad \text{ and } \quad \vert b \vert - \vert a \vert \leq \vert a - b \vert \\
\implies & \vert \vert a \vert - \vert b \vert \vert \leq \vert a - b \vert .
\end{align*}