WARNING
Homework
Math 555
Many of you have been making
(easily fixable since you have the right ideas) errors
in writing up your proofs that involve limits.
Basically, you have been writing
lim n&rarr&infin zn,
and assuming that the limit exists, before you
actually show the limit exists.
It is time to get it down right.
Below are some ideas that (hopefully) will help
prevent such troubles.
The idea behind the fix is to use that the lim-sup and lim-inf
of any sequence exist in the extended real numbers
R ∪ {+∞, -∞} .
So work with the lim-inf & lim-sup (keeping in mind that might be
+/- ∞, but atleast they exist)
instead of working with
lim n&rarr&infin zn
(which you don't even know if it exists).
So read the below.
You do not have to quote any of the Lemma/Theorems in
your proofs.
Just understand the ideas and use the ideas as
needed in your proofs.
A Basic Lemma
Let { zn } be a sequence from R
and M ε R.
Then the following are equivalent.
- (BL1)  
lim n&rarr&infin zn exists
(in R)
and lim n&rarr&infin zn = M
- (BL2)  
liminf n&rarr&infin zn
= limsup n&rarr&infin zn = M
- (BL3)  
For each ε > 0 we have that
M - ε < liminf n&rarr&infin zn
≤ limsup n&rarr&infin zn < M + &epsilon
Proof of Basic Lemma
Follows easily from the
following facts.
-
-∞ ≤ liminf n&rarr&infin zn
≤ limsup n&rarr&infin zn
≤ + ∞  
-
lim n&rarr&infin zn exists
(in R, ie. is finite)
if only only if
liminf n&rarr&infin zn
= limsup n&rarr&infin zn
(in R, ie. are finite) ,
in which case,
lim n&rarr&infin zn =
liminf n&rarr&infin zn
= limsup n&rarr&infin zn .
All done ♦
Squeeze Theorem (utility grade)
Let { zn } be a sequence from R
and M ε R.
If there exists squences
{ an } and
{ bn } from R
and N0 ε N
such that
( ∀ n≥ N0)
   
[ an ≤ zn ≤ bn ]
and
M = lim n&rarr&infin an
   and    
lim n&rarr&infin bn = M
then the limit, as n&rarr&infin, of
{ zn } exists and
lim n&rarr&infin zn = M    .
Proof of Squeeze Theorem (utility grade)
Theorem 2.2.4 from textbook. ♦
Squeeze Theorem (working version)
Let { zn } be a sequence from R
and M ε R.
If,
for each ε >0 ,
there exists squences
{ anε } and
{ bnε } from R
and Nε ε N
such that
( ∀ n≥ Nε)
   
[ anε
≤ zn ≤ bnε ]
and
M - ε ≤
liminf n&rarr&infin an ε
   and    
limsup n&rarr&infin bnε
≤ M + ε
then the limit, as n&rarr&infin, of
{ zn } exists and
lim n&rarr&infin zn = M    .
Proof of Squeeze Theorem (working version)
Let the givens (and the If) be given.
Then, for each ε >0 ,
liminf n&rarr&infin an ε
≤
liminf n&rarr&infin zn
≤
limsup n&rarr&infin zn
≤
limsup n&rarr&infin bn ε
and so
M- ε
≤
liminf n&rarr&infin zn
≤
limsup n&rarr&infin zn
≤
M + ε   .
Since ε >0 was arbitrary,
liminf n&rarr&infin zn
= M =
limsup n&rarr&infin zn   .
All done ♦
Findable from URL:
http://www.math.sc.edu/~girardi/