Limsup, liminf simply explained

**gotmejerry** · January 14th 2012, 11:19 AM

I don't really understand these and why are they important. Can someone explain me these concepts?
Thanks!

**JakeBarnes** · January 16th 2012, 06:33 AM

There are a few equivalent ways to define these concepts. Let $\{a_n\}_{n=1}^{\infty}$ be a sequence of real numbers. For each positive integer n, define the number $b_n := \sup \{a_k : k \ge n\}$ . Then we define $\limsup_{n \to \infty} a_n := \lim_{n \to \infty} b_n$ .

There are a few immediate consequences of this definition:

1. If $m \le n$ then $\{a_k : k \ge n\} \subset \{a_k : k \ge m\}$ . So $b_m \ge b_n$ , hence the sequence $\{b_n\}_{n=1}^{\infty}$ is a decreasing sequence. So $\lim_{n \to \infty} b_n$ always exists, since a monotonic sequence of real numbers always has a limit in the extended reals.

Similarly, we can define $c_n := \inf \{a_k : k \ge n\}$ and defining $\liminf_{n \to \infty} a_n := \lim_{n \to \infty} c_n$ . Then a similar analysis shows that the sequence $\{c_n\}_{n=1}^{\infty}$ is a monotonically increasing sequence, and so its limit always exists in the extended reals.

2. $\lim_{n \to \infty} a_n$ exists if and only if $\limsup_{n \to \infty} a_n = \liminf_{n \to \infty} a_n$ ; in which case, the limit is the same.

3. $\liminf_{n \to \infty} a_n \le \limsup_{n \to \infty} a_n$

4. If $B = \limsup_{n \to \infty} a_n$ and $C = \liminf_{n \to \infty} a_n$ , then B and C satisfy the following:

$a_n > B$ for only finitely many n and $a_n < C$ for only finitely many n.

5. There exist subsequences of $\{a_n\}_{n=1}^{\infty}$ which converge to B and C.

Now, why do we care? For starters, the limsup and liminf always exist. Compare this with the limit, which need not exist for any particular sequence. For example, the sequence $a_n = (-1)^n$ has no limit. The limsup and liminf provide tools which can be used to establish some properties for the sequence.

An elementary example of this is the root test from the analysis of infinite sums.

Root test: Given a sequence $\{a_n\}_{n=1}^{\infty}$ of real numbers, set $A := \limsup_{n \to \infty} (|a_n|)^{\frac{1}{n}}$

Then:

If A < 1, the sum converges
If A > 1, the sum diverges
If A = 1, there is no conclusion.

Now note in the statement of the theorem that there's no question of the existence of A. You might have encountered this theorem once before with the limsup replaced by an ordinary limit. The same result holds, so long as the limit exists. By extending the statement to deal with a limsup, there is no need to assume the stronger hypothesis that the limit exists.

Analysis is often concerned with estimations and approximations. The limsup and liminf provide a convenient method for constructing bounds on the end behavior of the sequence.

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