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Let $\{a_n\}$ be a sequence of real numbers bounded from above, $A\in \Bbb R$.

Given any $\epsilon>0$,

a)$\exists n_0 \in \Bbb N$ such that $ a_n<A+\epsilon$ for all $n\ge n_0$.

b)$\exists k\ge n_0$ such that $a_k>A-\epsilon$.

If the sequence satisfies the above two properties, show that $A=\limsup_\limits{n\to\infty}a_n$.

I know the definition of the limit superior as:$\limsup a_n = \inf_{\forall m} \sup_{n \ge m} a_n$. Also, if ($a_n$) is a real sequence bounded from above. Let $S :=$ {$t \in \Bbb R:$ $t$ is the limit of a convergent subsequence of ($a_n$) }. Then $A = sup S$. I've proved the opposite direction (i.e. Given $A=\limsup_{n\to\infty}a_n$, then it has the following two properties), but stuck on trying to prove the two properties imply A. Could someone provide a precise proof of this please? Thanks.

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  • $\begingroup$ Check out the \limits I added to the title, it makes the expression look cooler. $\endgroup$ – YoTengoUnLCD Feb 4 '16 at 3:34
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For any $\epsilon >0$ there exists $n_0 \in \mathbb{N}$ such that for all $n \geqslant n_0$ we have $a_n < A + \epsilon$ and $A- \epsilon < a_k$ for some $k \geqslant n_0$.

Hence,

$$A - \epsilon \leqslant \sup_{n \geqslant n_0}a_n \leqslant A + \epsilon.$$

For any $m > n_0$ we have $\sup_{n \geqslant m}a_n \leqslant \sup_{n \geqslant n_0}a_n$ since $[m,\infty) \subset [n_0,\infty).$ Also there exists $k > m$ such that $A-\epsilon < a_k$.

Whence, it follows that

$$A - \epsilon \leqslant \sup_{n \geqslant m}a_n \leqslant A + \epsilon.$$

By definition, $\limsup_{n \to \infty}a_n = \inf_{m} \sup_{n \geqslant m}a_n$ and, since $\sup_{n \geqslant m}a_n$ is decreasing and bounded below, $\limsup_{n \to \infty}a_n= \lim_{m \to \infty}\sup_{n \geqslant m}a_n.$

Therefore, we have for any $\epsilon > 0$

$$A - \epsilon \leqslant \limsup_{n \to \infty}a_n \leqslant A + \epsilon.$$

Now you can reach the conclusion.

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  • $\begingroup$ Can I say that hence $A-\epsilon \le lim sup a_n \le A+\epsilon$, and since $\epsilon$ is arbitrary, this can only happen if lim sup $a_n$ = $A$. $\endgroup$ – user57891 Feb 4 '16 at 3:21
  • $\begingroup$ That is the final step, but there are some intermediate steps. I can expand above if you don't see it. $\endgroup$ – RRL Feb 4 '16 at 3:24
  • $\begingroup$ If you could prove it in details that would be great, thanks! $\endgroup$ – user57891 Feb 4 '16 at 3:27

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