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.