Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 8 Exercise 8.10

This part actually has been shown in Exercise 8.9.

Solution: Suppose $\lim s_n=s$. Then we have $s – a>0$.

Let $\epsilon =\dfrac{s-a}{2}$. Then there exists $N>0$ such that if $n>N$ then

$$

|s_n-s|<\epsilon\Longrightarrow s_n – s>-\epsilon.

$$ If $n>N$, then we have

$$

s_n-s>-\epsilon.

$$ Therefore, we have

$$

s_n > s-\epsilon=s-\frac{s-a}{2}=\frac{a+s}{2}.

$$ for all $n>N$. Because $s > a$, we have

$$

\frac{a+s}{2} > a

$$ which implies that

$$

s_n > \frac{a+s}{2} > a

$$ for all $n>N$. We are done.