Answer
Let $\left\{ {{a_n}} \right\}$ be a convergent sequence of integers with limit $L$, then there exists a number $M$ such that ${a_n} = L$ for all $n \ge M$.
Work Step by Step
If $\left\{ {{a_n}} \right\}$ is a convergent sequence of integers with limit $L$, then $\mathop {\lim }\limits_{n \to \infty } {a_n} = L$.
Thus, by the limit definition: for every $\varepsilon > 0$, there is a number $M$ such that $\left| {{a_n} - L} \right| < \varepsilon $ for all $n>M$. We choose $\varepsilon = \frac{1}{2} > 0$. Then, there exist a number $M$ such that
$\left| {{a_n} - L} \right| < \frac{1}{2}$,
$ - \frac{1}{2} < {a_n} - L < \frac{1}{2}$,
$L - \frac{1}{2} < {a_n} < L + \frac{1}{2}$.
Since $\left\{ {{a_n}} \right\}$ is a sequence of integers, the last inequality says that ${a_n} = L$ for all $n \ge M$.
