1. ## Convergent Sequence Inequality

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

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

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

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

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

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

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