# What is limit superior and limit inferior?

I've looked at the Wikipedia article, but it seems like gibberish. The only thing I was able to pick out of it was the concept of infimum (greatest lower bound) and supremum (least upper bound), as I had learned them previously in an intro discrete math course.

The limit inferior of a sequence ($x_n$) is defined by

$\displaystyle\liminf_{n\to\infty}x_n := \lim_{n\to\infty}\Big(\inf_{m\geq n}x_m\Big)$

or

$\displaystyle\liminf_{n\to\infty}x_n := \sup_{n\geq 0}\,\inf_{m\geq n}x_m=\sup\{\,\inf\{\,x_m:m\geq n\,\}:n\geq 0\,\}.$

Similarly, the limit superior of ($x_n$) is defined by

$\displaystyle\limsup_{n\to\infty}x_n := \lim_{n\to\infty}\Big(\sup_{m\geq n}x_m\Big)$

or

$\displaystyle\limsup_{n\to\infty}x_n := \inf_{n\geq 0}\,\sup_{m\geq n}x_m=\inf\{\,\sup\{\,x_m:m\geq n\,\}:n\geq 0\,\}.$

Can anybody provide any examples of its use, and why it's used in that context?

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– Amzoti Jun 17 '13 at 17:42
An intuition is that $\limsup x_n$ is the maximum limit possible for some convergent sub-sequence of $x_n$ (including possibly $+\infty$.) So $x_n=(-1)^n$ has limit superior $1$ because there is s sub-sequence that coverges to $1$ and no sub-sequence converges to any higher value than $1$. – Thomas Andrews Jun 17 '13 at 17:42
@ThomasAndrews: Do you know how to prove the statement that you quoted? – Idonknow Jan 19 '14 at 10:41
ahh was gibberish to me too intially, but after looking at that first diagram on [wikipedia page] and its description (en.wikipedia.org/wiki/Limit_superior_and_limit_inferior), I can relate to what above definitions means. Future readers, check out that diagram and description and try relating. (Description can only be found on topic page, not on image page) – PardonMeForMySuperPoorMaths May 2 at 19:15

A very prominent application of $\limsup$ is the Cauchy-Hadamard formula for the radius of convergence: Given a power series $\sum_{n=0}^\infty a_n x^n$, its radius f convergence $R$ can be obtained from $$\frac1R=\limsup_{n\to\infty}\sqrt[n]{|a_n|}.$$ Without going into details why that is so, let's ask:

• Why not the $\lim$? Because it may not even exist (e.g. if $a_n=1+(-1)^n$).

• Why not the $\sup$? Because a single large $|a_n|$ would then spoil the value whereas a single summand $a_nx^n$ does not influence the convergence.

Nevertheless, if $L:=\lim_{n\to\infty}\sqrt[n]{|a_n|}$ happens to exist, we see that the power series is dominated by $\sum |L' x|^n$ for any $L'>L$, and this converges (to $\frac1{1-|L'x|}$) provided $|L'x|<1$, and by suitable choice of $L'$ we obtain convergence whenever $|Lx|<1$. But if the limit does not exist, we must use the $\limsup$, so we have a sub-sequence converging to some $L$ and can choose the subsequence so large that all other terms $\sqrt[n]{|a_n|}$ are less than $L$. These small terms don't hurt the convergence if $|xL|<1$. And yet, since our subsequence has infinitely many terms, this is enough to spoil convergence when $|xL|>1$.

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Ah. Well, I'm guessing that it's just a bit ahead of my time then. I'm starting power series and taylor series shortly. Maybe I'll see it there. – agent154 Jun 17 '13 at 18:00

I've found that some students have difficulty understanding the usual definitions of limit superior and inferior because these definitions combine the notions of limits, of suprema, and of infima, all of which the student may have learned only recently and not fully internalized. For such students, I like to give the following alternative definitions, equivalent to the usual ones but not containing the words "limit", "supremum", and "infimum". (Nor are there absolute values or visible $\varepsilon$'s.)

A number $t$ is the limit superior of a sequence $\langle a_n\rangle$ if the following two conditions are both satisfied:

• For every $s<t$ we have $s<a_n$ for infinitely many $n$'s.

• For every $s>t$ we have $s<a_n$ for only finitely many $n$'s (possibly none).

Similarly, a number $t$ is the limit inferior of a sequence $\langle a_n\rangle$ if the following two conditions are both satisfied:

• For every $s>t$ we have $s>a_n$ for infinitely many $n$'s.

• For every $s<t$ we have $s>a_n$ for only finitely many $n$'s (possibly none).

Two additional remarks may be useful:

1. The definition of lim inf is gotten from the definition of lim sup by simply reversing all inequalities.

2. The definitions can be easily extended to $\pm\infty$ in place of numbers $t$. Just adopt the convention that, even then, $s$ refers to actual numbers, all of which are $>-\infty$ and $<+\infty$.

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The following are answers of mine which use $\liminf$ and $\limsup$ [1] [2] [3] [4]

Maybe this helps.

DEF Given a sequence of real numbers $\langle a_n\rangle$, we say that $\ell \in \Bbb R^*=\Bbb R\cup\{+\infty,-\infty\}$ is a limit point of the sequence if there exists a subsequence $\langle a_{n_k}\rangle$ of $\langle a_n\rangle$ such that $$\lim_{k\to\infty}a_{n_k}=\ell$$

Now let's prove the

PROP Fix a sequence of real numbers $\langle a_n\rangle$, and define $$\mathscr L=\{x\in\Bbb R^*:x\text{ is a limit point of }\langle a_n\rangle\}$$ Then $\mathscr L$ is nonempty for any choice of $\langle a_n\rangle$.

P First suppose $\langle a_n\rangle$ is bounded. By Bolzano Weierstrass, there exists a convergent subsequence $\langle a_{n_k}\rangle$ of $\langle a_n\rangle$ such that $\lim\limits_{k\to\infty}a_{n_k}=\ell$ for some $\ell\in\Bbb R$. Thus $\ell\in\mathscr L$. Now assume $\langle a_n\rangle$ is unbounded. We can assume it is unbounded from above. Then, by definition, for each $k\in \Bbb N$ there exists $n_k$ such that $a_{n_k}\geq k$. It follows $\lim\limits_{k\to\infty}a_{n_k}=+\infty$, so $+\infty\in\mathscr L$.

Assume from now on the sequence is bounded.

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 $$A_n=\left[\underline{a_n},\overline{a_n}\right]$$

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

DEF For each sequence $\langle a_n\rangle$, define the intersection $$\bigcap_{n\in \Bbb N}A_n=[\zeta,\eta]$$This is nonempty courtesy of Cantor's intersection theorem.

Observe that $\zeta=\lim\limits_{n\to\infty} \underline{a_n}$ and $\eta=\lim\limits_{n\to\infty} \overline{a_n}$ 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 $\ell \in [\zeta,\eta]$. That is $\mathscr L\subseteq [\zeta,\eta]$.

$2.$ $\eta,\zeta$ are limit points of $\langle a_n\rangle$, thus conclude that $\eta,\zeta$ are the smallest and largest limit points of $\langle a_n\rangle$. Thus $\zeta=\sup\mathscr L=\max \mathscr L\; ,\; \eta=\inf\mathscr L=\min \mathscr L$.

$3.$ 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 $\lim\limits_{n\to\infty} a_n=p$, all subsequences converge to $p$, so the interval $[\zeta,\eta]$ degenerates to the single point $p=\eta=\zeta$.

NOTE If the sequence is unbounded from above (resp. below) then $$\limsup_{n\to\infty}a_n=+\infty\;\;\left( \liminf_{n\to\infty}a_n=-\infty\right)$$

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you've answered this question beautifully!! However, I haven't been able to fill in the proofs you've left to the reader. So can you please also supply these proofs? – Saaqib Mahmuud Jul 1 '15 at 20:38