Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 8 Exercise 8.6
Solution:
Part a
We first show that if $\lim s_n=0$ then $\lim |s_n|=0$.
Let $\epsilon>0$. Since $\lim s_n=0$, there exists $N>0$ such that
$$
|s_n-0|=|s_n|<\epsilon
$$ for all $n>N$. Therefore, $n>N$ implies that
$$
||s_n|-0|=|s_n|<\epsilon
$$ as desired. Thus $\lim |s_n|=0$.
Then we show that if $\lim |s_n|=0$ then $\lim s_n=0$.
Let $\epsilon>0$. Since $\lim |s_n|=0$, there exists $N>0$ such that
$$
||s_n|-0|=|s_n|<\epsilon
$$ for all $n>N$. Therefore, $n>N$ implies that
$$
|s_n-0|=|s_n|<\epsilon
$$ as desired. Thus $\lim s_n=0$.
Part b
Clearly, $|s_n|=1$ and hence we have $\lim|s_n|=1$ because $||s_n|-1|=0$ for all $n$.
However, $\lim s_n$ does not exist as you may see from Example 4.