The question I want to answer is : If ($a_n$) is a converging sequence, and for all n ∈ Z +, $a_n$ ≤ 1, then $lim_{n→∞}a_n$ ≤ 1. Here is my proof for this.
Proof: Suppose ($a_n$) is converges to some limit a, and for all n ∈ Z +, $a_n$ ≤ 1, and further suppose, absurdly, that $a>1$. Then $a-1>0$ and so we set $\epsilon=a-1 >0$. Then there exists $N_1\in$ N such that for all $n \in $ N, if $n>N_1$, then |$a_n−a$|< $\epsilon=a-1$. This implies that $a-a_n ≤ |a_n−a|$ which then implies that $a-a_n < a-1$ and thus $a_n>1$, which contradicts the assumption that $a_n$ ≤ 1. $\Box$
The proof looks fine!
An alternative approach is the following: the interval $I:=(-\infty, 1]$ is closed. Since $(a_n) \subset I$ and $a_n \to a$, it holds $a\in I$.