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This problem is purely for my own benefit, so I'd appreciate it if you offer help but don't spoil the proof for me. I've worked out the following solution, but I want to make sure that my reasoning doesn't have any holes in it:

Suppose we have the sequence $\{a_n\}$, and $\lim_{n \rightarrow \infty}\{a_n\}=L$. Then given $\epsilon>0$, we can choose $N \in \mathbb{N}$ such that $$\left| a_n-L \right|<\frac{\epsilon}{2}$$ for every term $a_n \in \{a_n\}$ such that $n>N$. Now for $\inf_{n>N}\{a_n\}$, (that is, to my understanding, the infimum of the sequence with the first N terms truncated) we must have some $a_{low} \in \{a_n|n>N\}$ such that $$a_{low}-\inf_{n>N}\{a_n\}<\frac{\epsilon}{2}$$ because otherwise $\inf_{n>N}\{a_n\}-\frac{\epsilon}{2}$ would be a greater lower bound for $\{a_n|n>N\}$. Similar reasoning demonstrates that we must have some $a_{high}$ such that $$\sup_{n>N}\{a_n\}-a_{high}<\frac{\epsilon}{2}$$ Now since $a_{high} \in \{a_n|n>N\}$, we must have $$-\epsilon<a_{high}-L<\epsilon$$ and, adding this inequality to the above one gives $$-\epsilon<\sup_{n>N}\{a_n\}-L<\epsilon$$ for $n>N$, and therefore $\lim_{N \rightarrow \infty}\sup_{n>N}\{a_n\}=L$. Likewise, since $a_{low} \in \{a_n|n>N\}$ we must have $$\left|a_{low}-L\right|=\left|L-a_{low}\right|<\frac{\epsilon}{2}$$ So $$-\frac{\epsilon}{2}<L-a_{low}<\frac{\epsilon}{2}$$ and adding this gives to the second inequality gives $$-\epsilon<L-\inf_{n>N}\{a_n\}<\epsilon$$ for $n>N$. So we have $\lim_{N \rightarrow \infty}\inf_{n>N}\{a_n\}=\lim_{N \rightarrow \infty}\sup_{n>N}\{a_n\}=L$. Now conversely assume that $\lim \inf=\lim \sup$. We have $$\inf_{n>N}\{a_n\} \leq a_n \leq \sup_{n>N}\{a_n\}$$ for $a_n \in \{a_n|n>N\}$. Since the limits of the left and right sides of the inequality are equal, the limit of the middle exists and equals that of the other two by the squeeze theorem (I've only seen the squeeze theorem proven for functions, but I believe it applies here since sequences are just functions with domains on the naturals). Therefore the sequence $\{a_n\}$ converges iff its $\lim \inf$ equals its $\lim \sup$ (and, as a corollary, the limit of the sequence always equals that of the infima and suprema).

Please tell me if you see any holes or things I could've done better, as I've made many stupid mistakes in proofs before. Thanks for your help!

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  • $\begingroup$ If you don't consider infinite limits as limits, you might want to add that your sequence is bounded. $\endgroup$ – Julien Mar 9 '13 at 20:04
  • $\begingroup$ @julien What do you mean I don't consider infinite limits as limits? $\endgroup$ – Liam Mar 9 '13 at 20:15
  • $\begingroup$ I mean, if $\lim x_n=+\infty$, some people say that the sequence does not converge. $\endgroup$ – Julien Mar 9 '13 at 20:18
  • $\begingroup$ I edited, tell me if it helps. Your proofs are fine. $\endgroup$ – Pedro Tamaroff Mar 9 '13 at 22:19
  • $\begingroup$ Related question: math.stackexchange.com/questions/432163/… $\endgroup$ – Martin Sleziak Jan 24 '16 at 22:19
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I'm sorry I don't really follow your argument. So I'll let others tell you if it's fine. Here is how I would do this.

Let $(x_n)$ be a bounded sequence in $\mathbb{R}$.

Then the set $C$ all limits of converging subsequences is non empty and $$ C\subseteq [\liminf x_n,\limsup x_n]. $$

If $\liminf x_n=\limsup x_n$, it follows that $$ C=\{x\}. $$ In other words, every converging subsequence of $(x_n)$ converges to the same $x$.

Now assume that $x_n$ does not converge to $x$ and construct a subsequence which converges to a point other than $x$.

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  • $\begingroup$ Why the downvote? What is wrong with this argument? Tell me and I'll fix it. $\endgroup$ – Julien Mar 9 '13 at 21:16
  • $\begingroup$ @JasperLoy You're right. I should stop asking... $\endgroup$ – Julien Mar 9 '13 at 21:39
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To directly answer your question. Your proof that $\ell =\limsup=\liminf\implies \lim a_n=\ell$ is fine. The squeeze theorem applies to sequences to.

Conversely, suppose that $\lim a_n=\ell $ is finite. Since $\langle a_k:k\geq n\rangle $ is bounded, $\inf$ and $\sup$ exist. As you claim, for each $\epsilon >0$ there exists an element $a_m$ from $ \langle a_k:k\geq n\rangle $ such that $\limsup a_n-a_m<\epsilon$. Similarily, there exists $a_w$ from $ \langle a_k:k\geq n\rangle $ such that $a_w-\liminf a_n<\epsilon$. You can use the triangle inequality to conclude that $\limsup=\lim=\liminf$ by noting that the $M$ you choose to get $a_n$ near to $\lim a_n$ can be used as a "parameter" in $ \langle a_k:k\geq M\rangle $


DSC This is not an answer, but rather, a long comment/help.

There is an interesting definition of $\limsup$ and $\liminf$ you might find useful.

DEF Suppose we're given a sequence $\langle a_n\rangle$ of real numbers. Then we say $\ell$ is a limit point of $\langle a_n\rangle$ if there exists some subsequence $\langle a_{n_k}\rangle$ of $\langle a_n\rangle$ which converges to $\ell$.

DEF Let $\langle a_k\rangle$ be a sequence in $\Bbb R$. We define for each $n\in \Bbb N$ the associated sequences $$\overline{a}_n=\sup \langle a_k:k\geq n\rangle$$ $$\underline{a}_n=\inf \langle a_k:k\geq n\rangle$$ and subsequently the closed intervals $$A_n=[\underline{a}_n,\overline{a}_n]$$

Observe that for each $n$, $$A_{n+1}\subseteq A_n$$

DEF For each sequence $\langle a_n\rangle$, define the intersection $$\bigcap_{n\in \Bbb N}A_n=[\zeta,\eta]$$

Or more informally "$=\lim A_n$".

Observe that $$\zeta=\lim \underline{a}_n$$ and $$\eta=\lim \overline{a}_n$$ are just the $\limsup$ and $\liminf$ of $\langle a_n\rangle $.

Prove

$1.$ If $\ell$ is a limit point of $\langle a_n\rangle $, then $\ell \in [\zeta,\eta]$.

$2.$ $\eta,\zeta$ are limit points of $\langle a_n\rangle $, thus conclude that $\eta,\zeta$ are the smallest and largest limit points of $\langle a_n\rangle $

$3.$ Observe that if $\zeta=\eta$, the interval degenerates to a single point $p=\zeta=\eta$, which means that the trivial subsequence $\langle a_n\rangle $ converges to $p=\zeta=\eta$. Conversely, if $\lim a_n=p$, all subsequences converge to $p$, so the interval $[\zeta,\eta]$ degenerates to the single point $p=\eta=\zeta$.

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  • $\begingroup$ I think we share the same downvoter, given that our approaches are similar. I'll be happy to know what he/she finds wrong about it. +1. $\endgroup$ – Julien Mar 9 '13 at 21:18
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    $\begingroup$ @julien Totally $\endgroup$ – Pedro Tamaroff Mar 9 '13 at 21:58
  • $\begingroup$ Nice drawing, +1. $\endgroup$ – Julien Mar 9 '13 at 22:01
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(I)Suppose we have the sequence $\{a_n\}$, and $\lim_{n \to \infty}\{a_n\}=L$. (The curly braces in the limit expression is redundant.) Then given $\epsilon>0$, we can choose $N \in \mathbb{N}$ such that $$ \left| a_n-L \right|<\frac{\epsilon}{2} $$ for every term a_n \in {a_n} such that $n>N$. Now for $\inf_{n>N}\{a_n\}$, (that is, to my understanding, the infimum of the sequence with the first N terms truncated) we must have some $a_{low} \in \{a_n|n>N\}$ such that $$ a_{low}-\inf_{n>N}\{a_n\}<\frac{\epsilon}{2} $$ because otherwise $\inf_{n>N}\{a_n\}-\frac{\epsilon}{2}$ (Typo here: the minus sign should be +) would be a greater lower bound for the set $\{a_n|n>N\}$.

Similar reasoning demonstrates that we must have some $a_{high}$ such that $$ \sup_{n>N}\{a_n\}-a_{high}<\frac{\epsilon}{2} $$ Now since $a_{high} \in \{a_n|n>N\}$, we must have $$ -\epsilon<a_{high}-L<\epsilon $$ (you should have $\epsilon/2$ in this inequality) and, adding this inequality to the above one gives $$ -\epsilon<\sup_{n>N}\{a_n\}-L<\epsilon\tag{*} $$ for $n>N$, (This is incorrect: the statement * does not depend on n at all.) and therefore $\lim_{N \to \infty}\sup_{n>N}\{a_n\}=L$. (This conclusion is logically incorrect: in order to make this conclusion, you need * to be true for all sufficiently large $N$.)

Likewise, since $a_{low} \in \{a_n|n>N\}$ we must have $$\left|a_{low}-L\right|=\left|L-a_{low}\right|<\frac{\epsilon}{2}$$ So $$-\frac{\epsilon}{2}<L-a_{low}<\frac{\epsilon}{2}$$ and adding this gives to the second inequality gives $$-\epsilon<L-\inf_{n>N}\{a_n\}<\epsilon$$ for $n>N$. So we have $\lim_{N \rightarrow \infty}\inf_{n>N}\{a_n\}=\lim_{N \rightarrow \infty}\sup_{n>N}\{a_n\}=L$.

(II)Now conversely assume that $\lim \inf=\lim \sup$. We have $$ \inf_{n>N}\{a_n\} \leq a_n \leq \sup_{n>N}\{a_n\} $$ for $a_n \in \{a_n|n>N\}$. (The small $n$ in the first and third terms are dummy variables while in the middle term it is not. If you want to use the squeeze theorem, them the inequality should be $$ \text{for each } N\quad \inf_{n>N}\{a_n\} \leq a_{N+1} \leq \sup_{n>N}\{a_n\} $$ and when taking the limit, you want $N\to\infty$.) Since the limits of the left and right sides of the inequality are equal, the limit of the middle exists and equals that of the other two by the squeeze theorem (I've only seen the squeeze theorem proven for functions, but I believe it applies here since sequences are just functions with domains on the naturals).

Therefore the sequence $\{a_n\}$ converges iff its $\lim \inf$ equals its $\lim \sup$ (and, as a corollary, the limit of the sequence always equals that of the infima and suprema).

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