1. ## Convergent Sequence Inequality

If $\displaystyle a_n\,$ is a convergent sequence and $\displaystyle \lim a_n >a\,$. Show that $\displaystyle a_n > a\,$ for some $\displaystyle n>N\,$.

My attempt was like this.
I used the limit limitation theorem*.

Assume $\displaystyle a_n\leq a$ for sufficiently large $\displaystyle n\,$
Then, $\displaystyle \lim a_n\leq a$

Thus, the statement $\displaystyle a_n\leq a$ for sufficiently large $\displaystyle n\,$ is false.
My difficutly is that we cannot conclude that $\displaystyle a_n>a$.
Because that cannot be the negation of the statement.
It may mean that, there is no such $\displaystyle N\,$ such
$\displaystyle a_n\leq a$ for $\displaystyle n>N$.
Meaning we can have,
$\displaystyle a_n>a$ for $\displaystyle n=N+1$.
$\displaystyle a_n\leq a$ for $\displaystyle n=N+2$.
$\displaystyle a_n > a$ for $\displaystyle n=N+3$.
And thus on....

*)If $\displaystyle a_n\,$ is a convergent sequence and $\displaystyle a_n\geq a\,$ for sufficiently large $\displaystyle n\,$ then $\displaystyle \lim a_n\geq a\,$.

2. For notation sake say $\displaystyle \left( {a_n } \right) \to L > a$.
Then we have:
$\displaystyle \begin{array}{rcl} \frac{{L - a}}{2} > 0\quad & \Rightarrow & \quad \left( {\exists K \in Z^ + } \right)\left[ {\left| {a_K - L} \right| < \frac{{L - a}}{2}} \right] \\ \quad & \Rightarrow & \quad - \frac{{L - a}}{2} < a_K - L \\ & \Rightarrow & \quad L - \frac{{L - a}}{2} < a_K \\ & \Rightarrow & \quad a < \frac{{L + a}}{2} < a_K \\ \end{array}$

3. Originally Posted by Plato
For notation sake say $\displaystyle \left( {a_n } \right) \to L > a$.
Then we have:
$\displaystyle \begin{array}{rcl} \frac{{L - a}}{2} > 0\quad & \Rightarrow & \quad \left( {\exists K \in Z^ + } \right)\left[ {\left| {a_K - L} \right| < \frac{{L - a}}{2}} \right] \\ \quad & \Rightarrow & \quad - \frac{{L - a}}{2} < a_K - L \\ & \Rightarrow & \quad L - \frac{{L - a}}{2} < a_K \\ & \Rightarrow & \quad a < \frac{{L + a}}{2} < a_K \\ \end{array}$
Thank you, excellent job.