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.

  • $\begingroup$ Do you really mean to ask if $|\lim_{n\to \infty} a_{n}| \leq \alpha$? That seems to better match the title of the post. (It doesn't actually matter, but technically...) $\endgroup$ – Quinn Culver May 27 '11 at 21:04
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    $\begingroup$ It's true. However, you cannot replace weak inequalities by strict ones. (Even if $|a_n| < \alpha$ for each $n$, we can guarantee only that $\lim |a_n| \le \alpha$.) $\endgroup$ – Sammy Black May 27 '11 at 21:06
  • $\begingroup$ @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? $\endgroup$ – ItsNotObvious May 27 '11 at 21:45

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 $(a_n)$.

  • $\begingroup$ Indeed, that's a much simpler proof. $\endgroup$ – ItsNotObvious May 27 '11 at 21:45

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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    $\begingroup$ Two comments: 1. When you define $a$, you include the inequality $\le \alpha$, which is your conclusion! 2. The strict inequalities in Eqn. 1, allow you to strengthen the final inequality to a strict $<$. (Ultimately, it doesn't matter.) $\endgroup$ – Sammy Black May 27 '11 at 21:07
  • $\begingroup$ 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 it $\endgroup$ – ItsNotObvious May 27 '11 at 21:35

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