Facts about limits of sequences

In this section we will go over some basic results about the limits of sequences. We start with
looking at how sequences interact with inequalities.
2.2.1 Limits and inequalities
A basic lemma about limits is the so called squeeze lemma. It allows us to show convergence of
sequences in difficult cases if we can find two other simpler convergent sequences that “squeeze”
the original sequence.
Lemma 2.2.1 (Squeeze lemma). Let fang, fbng, and fxng be sequences such that
an  xn  bn for all n 2 N:
Suppose that fang and fbng converge and
lim
n!¥
an = lim
n!¥
bn:
Then fxng converges and
lim
n!¥
xn = lim
n!¥
an = lim
n!¥
bn:
The intuitive idea of the proof is best illustrated on a picture, see Figure 2.1. If x is the limit of
an and bn, then if they are both within e=3 of x, then the distance between an and bn is at most 2e=3.
As xn is between an and bn it is at most 2e=3 from an. Since an is at most e=3 away from x, then xn
must be at most e away from x. Let us follow through on this intuition rigorously.
an x xn bn
Figure 2.1: Squeeze lemma in picture.
Proof. Let x := lim an = lim bn. Let e > 0 be given.
Find an M1 such that for all n  M1 we have that jan􀀀xj < e=3, and an M2 such that for all
n  M2 we have jbn􀀀xj < e=3. Set M := maxfM1;M2g. Suppose that n  M. We compute
jxn􀀀anj = xn􀀀an  bn􀀀an
= jbn􀀀x+x􀀀anj
 jbn􀀀xj+jx􀀀anj
<
e
3
+
e
3
=
2e
3
:
48 CHAPTER 2. SEQUENCES AND SERIES
Armed with this information we estimate
jxn􀀀xj = jxn􀀀x+an􀀀anj
 jxn􀀀anj+jan􀀀xj
<
2e
3
+
e
3
= e:
And we are done.
Example 2.2.2: A simple example of how to use the squeeze lemma is to compute limits of
sequences using limits that are already known. For example, suppose that we have the sequence
f 1
n
p
ng. Since
p
n  1 for all n 2 N we have
0 
1
n
p
n

1
n
:
for all n 2 N. We already know that lim1=n = 0. Hence, using the constant sequence f0g and the
sequence f1=ng in the squeeze lemma, we conclude that
lim
n!¥
1
n
p
n
= 0:
Limits also preserve inequalities.
Lemma 2.2.3. Let fxng and fyng be convergent sequences and
xn  yn;
for all n 2 N. Then
lim
n!¥
xn  lim
n!¥
yn:
Proof. Let x := lim xn and y := lim yn. Let e > 0 be given. Find an M1 such that for all n  M1 we
have jxn􀀀xj < e=2. Find an M2 such that for all n  M2 we have jyn􀀀yj < e=2. In particular, for
n  maxfM1;M2g we have x􀀀xn < e=2 and yn􀀀y < e=2. We add these inequalities to obtain
yn􀀀xn+x􀀀y < e; or yn􀀀xn < y􀀀x+e:
Since xn  yn we have 0  yn􀀀xn and hence
0 < y􀀀x+e; or 􀀀e < y􀀀x:
In other words, x􀀀y<e for all e >0. That means that x􀀀y0, as we have seen that a nonnegative
number less than any positive e is zero. Therefore x  y.
2.2. FACTS ABOUT LIMITS OF SEQUENCES 49
We give an easy corollary that can be proved using constant sequences and an application of
Lemma 2.2.3. The proof is left as an exercise.
Corollary 2.2.4.
i) Let fxng be a convergent sequence such that xn  0, then
lim
n!¥
xn  0:
ii) Let a;b 2 R and let fxng be a convergent sequence such that
a  xn  b;
for all n 2 N. Then
a  lim
n!¥
xn  b:
Note in Lemma 2.2.3 you cannot simply replace all the non-strict inequalities with strict
inequalities. For example, let xn := 􀀀1=n and yn := 1=n. Then xn < yn, xn < 0, and yn > 0 for all n.
However, these inequalities are not preserved by the limit operation as we have lim xn = lim yn = 0.
The moral of this example is that strict inequalities may become non-strict inequalities when limits
are applied. That is, if we know that xn < yn for all n, we can only conclude that
lim
n!¥
xn  lim
n!¥
yn:
This issue is a common source of errors.