Can it be shown that the limit of a bounded sequence is no greater than the bounding value? Let $(a_n)$ be a convergent sequence. Since $(a_n)$ converges it is bounded and therefore there exists a number $\alpha \geq 0$ such that $|a_n| \leq \alpha \; \forall \; n \in \mathbb{N}$. Is it true $\lim_{n \to \infty} |a_n| \leq \alpha$  ?
I believe it is true and my proposed answer to this question will attempt to confirm this belief.
 A: As a general tip, usually when I believe something is true, a proof by contradiction is in order. In this case if $a_{n} \to a$, but $a>\alpha$ then $a-\epsilon>\alpha$ for some $\epsilon>0$. By definition of convergence, there is some $a_N \in (a-\epsilon, a+\epsilon)$, but then $a_N > a-\epsilon >\alpha$, contrary to $\alpha$ being an upper bound for $(a_n)$.
A: Since $(a_n)$ converges, $(|a_n|)$ converges, and thus the limit of the sequence $|a_n|$ is defined. Let a = $\lim_{n \to \infty} |a_n|$. Then by definition of limit, for every $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that $n > N \implies |a_n - a| < \epsilon$ from which it follows that

(1)
$$
-\epsilon < a - a_n < \epsilon \; \forall \; n > N
$$
Since $(a_n)$ is bounded

(2)
$$
-\alpha \leq a_n \leq \alpha \; \forall \; n \in \mathbb{N}
$$
Combining (1) and (2),
$$
|a| < \alpha + \epsilon \; \forall \; \epsilon > 0
$$
Since this equation holds for all positive $\epsilon$ it must be the case that $|a| \leq \alpha$. Moroever, the limit of a nonnegative sequence must also be nonnegative and therefore $|a| = a$ and the claim is proved.
