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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.

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    @Quinn: no, the equation as written is what I intended to ask, but that was the best title I could come up with. Do you have a suggestion for an alternate title that would better reflect the question?2011-05-27

2 Answers 2

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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 $(a_n)$.

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    Indeed, that's a much simpler proof.2011-05-27
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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.

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    For comment 1: that was$a$typo which is now fixed. Thanks for pointing it out For comment 2, you're right and I'll fix it2011-05-27