Class notes Sept. 25
Some examples on limits of sequences
Exercise 2.3.3. (Squeeze Theorem):
lim xn  lim zn  l , then lim yn  l as well.
n 
n 
If
xn  yn  zn , for all n  N ,
and
if
n 
Proof: Let   0 . Since the sequences {xn } and {zn } converge, there exits n0  N such


that for all n  n0 , | xn  l | , and | zn  l | .
2
2
Now, by Theorem 2.3.3, lim( zn  xn )  0 , and moreover, zn  xn  0 for all n  N .
n 
Therefore, for   0 there exists n1  N such that for all n  n1 , zn  xn 

2
(we don’t
need in absolute value since the difference is positive).
Let’s then take n  max{n0 , n1} . We have
| yn  l || yn  xn |  | xn  l | zn  xn  | xn  l |

2


2
 ,
proving that lim yn  l .
n 
Exercise 2.3.4. (Uniqueness of the limit of a sequence) If a sequence an converges, then
its limit is unique.
Proof: Assume lim an  l1 and lim an  l2 , and suppose l1  l2 . Then either l1  l2 , or
n 
n 
l1  l2 . The two cases are similar so we will only analyze one of them. Suppose l1  l2 .
d
Let d  l2  l1  0 . Since lim an  l1 and lim an  l2 , for  
there must exist n0  N
n 
n 
4
such that | an  l1 |  , and | an  l2 |  for all n  n0 .
d
We have d  l2  l1 | l2  an |  | an  l1 | 2  , impossible! Therefore the limit has to be
2
unique.
Exercise 2.3.11. Césaro Means: If ( xn ) is a convergent sequence, then the sequence
x  x  ...  xn
given by the averages yn  1 2
converges to the same limit.
n
Proof: Denote lim xn  l . Here is the idea of the proof: since the sequence ( xn )
n 
converges, for a given   0 we will find a rank, say m  N such that for
n  m, | xn  l | 1 (1), where  1 will be defined shortly. We need to look at
n
| yn  l ||
x
j 1
n
j
 (x
j 1
j
 l)
 l ||
| . We will split the sum in the numerator into the sum up
n
n
to m 1 and the rest of the terms. We have:
m 1
n
| yn  l ||
 (x j  l)
j 1
n
|
| x j  l |
j 1
n
n

| x
j m
l |
(2)
n
n
| x
j
l |
(n  m  1)1
 1 (3).
n
n
Since ( xn ) converges, it must be bounded (see Theorem 2.3.2), therefore there exists
Now, according to (1), and the triangle inequality,
j m
j

M  0 , such that | xn | M for all n  N . Moreover, it follows that the limit is bounded
by M as well, so | l | M . Again, using the triangle inequality,
m 1
| x j  l |
m 1
 (| x
|  | l |)
2(m  1) M
(4).
n
n
n
2(m  1) M
 1 can be
So in order to prove that ( yn ) converges we need to show that
n
made arbitrarily small. But if we evaluate what we have in the above expression, M is a
fixed number, m will be again a fix number once we give ourselves   0 , but the
j 1

j 1
j

denominator n can get arbitrarily large. Therefore, if we define 1 
n0  N , such that for all n  n0 ,

2
, we can find
2(m  1) M

 1  (5). Putting together (3), (4) and (5),
n
2


  , for all n  n0 , proving that lim yn  l .
n 
2 2
Now let us look at the second part of the exercise: it basically says that if the Césaro
means of a sequence converge, the sequence must not necessarily converge. We will
show this by constructing a counterexample. Let xn  ( 1) n . We have:
the inequality in (2) becomes: | yn  l |

 1
x1  x2  ...  xn  , if n is odd
yn 
 n
n
 0, if n is even
Then lim yn  0 , but the sequence ( xn ) does not converge.
n 
1
.
4  xn
We can try to calculate a few terms to get a “feel” of the sequence. We have
1
3
x2  1, x3  , x4  . It looks like the terms are getting smaller and smaller. We see that
3
11
x2  x1 . We will conjecture that the sequence is decreasing, but before we prove it we
will show that it is bounded.
We prove that 0  xn  3, for all n  1 . This is obviously true for n  1 . Let us assume
Exercise 2.4.2. We are given the sequence defined by x1  3, xn 1 
that the inequality holds true for some n, that is, we have 0  xn  3 . Then
1
1
1
 xn 1 
 , and therefore 0  xn 1  3 . So we proved that for
4
4  xn 1
0  xn  3, for all n  1 .
1  4  xn  4, so
Next we show that the sequence is decreasing, xn1  xn , for all n  1 .
We already checked the statement for n  1 . Let us assume that it holds true for some n,
that is, we have xn1  xn .
1
We have xn  2 
. Since we proved that the sequence is bounded above by 3, we
4  xn 1
have
and
by
the
induction
assumption
4  xn 1  0 ,
1
1
xn 1  xn , therefore

 xn 1 , so xn 2  xn1 . It then follows that the
4  xn 1 4  xn
sequence is decreasing.
Now, by the Monotone Convergence Theorem (Theorem 2.4.2), the sequence converges,
so there must exist l  lim xn . Since the limit exists, we may take the limit in the
n 
1
. Solving for l in the previous equation we
4l
find two solutions, l1  2  3, and l2  2  3 . Now, both of them cannot be the limit of
recurrence relation, which gives us: l 
our sequence, and since we proved that 0  xn  3, for all n  1 , l1  2  3 cannot be the
limit of the sequence. Therefore lim xn  2  3 .
n 
Exercise 2.5.1 (Theorem 2.5.2) Any subsequence of a convergent sequence converges to
the same limit as the original sequence.
Proof:
Let ( xn ) be a convergent sequence, and let x  lim xn . Let ( xnk ) be a subsequence of
( xn ) . For any   0 there exists a rank N 0 such that for any n  N0 :| xn  x |  . Now,
since ( xnk ) is a subsequence the indices nk are positive integers, so there must exist an
index l  1 such that nl  N 0 (more precisely, define l  min{k : nk  N0 } , that is, nl is
the first index greater or equal to N 0 ). Then, for all k  l , | xnk  x |  , and therefore the
subsequence ( xnk ) has the same limit x.
Below we will provide a more detailed proof of the Bolzano-Weierstarss Theorem.
Theorem 2.5.5 (in Abbott) Every bounded sequence contains a convergent subsequence.
Proof:
Let ( xn ) be a bounded sequence. Therefore there exists M  0 :| xn | M for all n  1, or
M  xn  M , for all n  1 . Consider the two subintervals [ M , 0] and [0, M ] . At least
on of them must contain infinitely many terms of the sequence ( xn ) . Call that interval
I1  [a1 , b1 ] (this is just a notation; of course we either have a1  M , b1  0 or
a1  0, b1  M ). Now we define the first term in the subsequence to be the first in the
original sequence that belongs to the interval I1 , that is, let n1  min{n : xn  I1} . So we
have xn1  I1 .
Repeat the procedure for the interval I1 : consider the closed subintervals
a b a b
[a1 , 1 1 ],[ 1 1 , b1 ] . At least one of the subintervals contains infinitely many terms of
2
2
the sequence ( xn ) . Select the subinterval that does, and denote it I 2  [a2 , b2 ] . Define the
second term in the subsequence xn2  I 2 , where n2  min{n  n1 : xn  I 2 } .
We obtain in this way a recurrent definition of the intervals I k  [ak , bk ] satisfying
1
I k  I k 1 , and also bk  ak  (bk 1  ak 1 ) . Since b1  a1  M , it follows (by a simple
2
M
induction argument) that bk  ak  k 1 .
2
We also obtain the subsequence ( xnk ), ak  xnk  bk , k  1 . We claim that this is a
convergent subsequence.
Let us first prove that the family of closed nested intervals I1  I 2  ...  I k  ... has a
nonempty intersection, that is we prove there exists x :{x}  I k .
Indeed, we have the sequence ( ak ) of left end-points of the intervals, which is increasing
and bounded (for every k, M  ak  M , and the sequence (bk ) of right end-points of the
intervals, which is decreasing and bounded (for every k, M  bk  M ). Moreover,
M
bk  ak  k 1 
 0 . Therefore, by the Monotone Convergence Theorem (theorem
k 
2
2.4.2), both sequences have limits: there exist a  lim ak , b  lim bk , and a  b . But since
bk  ak  0 , it follows that a  b (details of this conclusion are shown in Lemma 1). So
we found a point x  a  b in the intersection of the sets. Lemma 1 will also show that
this is the only element of the intersection.
We now prove that the subsequence ( xnk ) converges to x. Let   0 . There exists k  1
1
1
  (indeed, just consider k  1  log 2 ). Then, for any
k 1
2

l  k :| xnl  x | bk  ak   , therefore lim xnk  x .
such that bk  ak 
QED
Lemma 1. Let I k  I k 1 , k  1 be the intervals constructed in Theorem 2.5.5. Then there
exists x :{x}  I k .
Proof: We proved in Theorem 2.5.5 that there exist a  lim ak  b  lim bk , where
I k  [ak , bk ] . If a  b it follows that ak  a  b  bk , for all k  1 , so [a, b]  I k . Since
1
1
bk  ak  k 1  0 , there must exist some k  1 such that k 1  b  a , but this
2
2
contradicts the fact that ak  a  b  bk , for all k  1 . Therefore a  b and the intersection
I k contains a single point.
QED