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Limit Superior and Limit Inferior

```Date: 03/27/2005 at 17:18:40
From: Joshua
Subject: lim sup and lim inf

Hello, I'm reading a real analysis text and I'm having trouble
understanding completely what lim sup and lim inf are.

First of all, is "lim sup" supposed to be looked at as a single
"operation"?  Second, I don't really understand the notation and how
the mechanism works according to the notations.  You can look here for
what I have in mind:

http://en.wikipedia.org/wiki/Limit_inferior

Lastly, I just don't understand the concept at all.  Can you first
give me an intuitive understanding and then a more formal way of
explaining it?  Thank you very much!

```

```
Date: 03/30/2005 at 20:20:13
From: Doctor Fenton
Subject: Re: lim sup and lim inf

Hi Joshua,

Thanks for writing to Dr. Math.  Yes, "lim sup" is a single process
(and the same is true of lim inf); it is related to the usual limit
process.

If you have a bounded sequence x_1,x_2,x_3,..., then you can form a
new sequence y_n by defining

y_n = sup{x_k: k >= n}.

I assume that you understand what the least upper bound or supremum of
a set is: it is the smallest number which is at least as large as
every element of the set.  (In particular, if the set has a largest
element, it is the supremum of the set.)

Notice that

y_(n+1) = sup {x_(n+1),x_(n+2),x_(n+3),...}

is the least upper bound of a smaller set than

y_n = sup {x_n,x_(n+1),x_(n+2),...}

so y_(n+1) <= y_n , and the sequence y_n is a decreasing sequence
which is bounded below (by any element of the sequence).  Since the
y_n sequence is decreasing and bounded below, it has a limit, call it
L, and

L = lim sup x_n =  lim y_n  ,
n->oo          n->oo

(or, since y_n is decreasing, the limit of the sequence must also be
the infimum or greatest lower bound of the set {y_n}, and

L = lim sup x_n =  inf y_n  . )
n->oo           n>0

With this latter characterization of the lim sup as the greatest lower
bound of the y_n, any larger number cannot be a lower bound for the
y_n, so if E is any positive number, L+E is not a lower bound for the
y_n, and there must be some y_N < L+E, so that

L <= y_N < L+E .

Since y_N is the least upper bound of {x_N,x_(N+1),x_(N+2),..}, then
y_N - E cannot be an upper bound for this set, which means that there
must be an x_m with m > N such that

y_N - E < x_m ,

so that

L - E <= y_N - E < x_m < y_N < L + E .

Since E is arbitrary (i.e. can be as small as we wish), this shows
that there is a subsequence of the x_n 's (formed from the x_m terms)
which converges to L.

For each E > 0, there can only be a finite number of x_n 's which are
larger than L+E, so no subsequence of x_n can converge to a larger
value than L.

So, in summary, the lim sup of the x_n is the largest number L such
that there is a subsequence of x_n's which converge to L.

A similar analysis leads to the conclusion that the lim inf x_n is the
smallest number for which there is a subsequence of the x_n which
converges to it.

If  lim x_n = L  exists, then  lim sup x_n = lim inf x_n = L .
n->oo                        n->oo        n->oo

If the two are different, then in a sense they measure how much the
sequence fails to have a limit: there is one subsequence which
converges to lim sup x_n, and another which converges to lim inf x_n .

As an example, let x_n = (-1)^n * (n+1)/n , so

x_1 = -2, x_2 = 3/2, x_3 = -4/3, x_4 = 5/4 , x_5 = -6/5, x_6 = 7/6

and so on.  The odd terms x_1,x_3,x_5,... increase to -1, while the
even terms x_2,x_4,x_6,... decrease to +1, so

lim sup (-1)^2 (n+1)/n = +1    and   lim inf (-1)^n (n+1)/n = -1 .

If you have any questions, please write back and I will try to explain
further.

- Doctor Fenton, The Math Forum
http://mathforum.org/dr.math/
```
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