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If the limit is greater than a number, so are all but finitely many of numbers in this sequence

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

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