4.4 The Limit Of A Sequence
The concept of the limit of a sequence was defined precisely by
The following is the precise definition of what is meant by the limit of a sequence.
Definition 4.4.1 A sequence
n=1∞ converges to a,
if and only if for every ε > 0 there exists nε such that whenever n ≥ nε ,
Here a and an are assumed to be complex numbers but the same definition holds more
In words the definition says that given any measure of closeness ε, the terms of the
sequence are eventually this close to a. Here, the word “eventually” refers to n being
sufficiently large. The above definition is always the definition of what is meant by the limit of
a sequence. If the an are complex numbers or later on vectors, the definition remains the
same. If an = xn + iyn and a = x + iy,
Recall the way you
measure distance between two complex numbers.
Theorem 4.4.2 If limn→∞an = a and limn→∞an = â then â = a.
Proof:Suppose â≠a. Then let 0 < ε <
2 in the definition of the limit. It follows
there exists nε
such that if n ≥ nε,
Therefore, for such
a contradiction. ■
Then it seems clear that
In fact, this is true from the definition. Let ε > 0 be given. Let nε ≥
it follows that n2 + 1 > ε−1 and so
is this large.
Note the definition was of no use in finding a candidate for the limit. This had to be
produced based on other considerations. The definition is for verifying beyond any doubt that
something is the limit. It is also what must be referred to in establishing theorems which are
good for finding limits.
Example 4.4.4 Let an = n2
Then in this case limn→∞an does not exist.
Example 4.4.5 Let an =
In this case, limn→∞
does not exist. This follows from the definition. Let ε
If there exists a limit, l,
then eventually, for all n
= 2 and so,
which cannot hold. Therefore, there can be no limit for this sequence.
Theorem 4.4.6 Suppose
are sequences and that
Also suppose x and y are in F. Then
Proof: The first of these claims is left for you to do. To do the second, let ε > 0 be given
and choose n1 such that if n ≥ n1 then
Then for such n, the triangle inequality implies
Now let n2
be large enough that for n ≥ n2,
Such a number exists because of the definition of limit. Therefore, let
For n ≥ nε,
This proves 4.3
. Next consider 4.4
Let ε > 0 be given and let n1 be so large that whenever n ≥ n1,
Thus for such n,
Now choose n2 so large that if n ≥ n2, then
Letting nε > max
it follows that for n ≥ nε,
Another very useful theorem for finding limits is the squeezing theorem.
Theorem 4.4.7 Suppose limn→∞an = a = limn→∞bn and an ≤ cn ≤ bn for
all n large enough. Then limn→∞cn = a.
Proof: Let ε > 0 be given and let n1 be large enough that if n ≥ n1,
Then for such n,
The reason for this is that if cn ≥ a, then
because bn ≥ cn. On the other hand, if cn ≤ a, then
As an example, consider the following.
Example 4.4.8 Let
and let bn =
, and an
. Then you may easily show that
Since an ≤ cn ≤ bn, it follows limn→∞cn = 0 also.
Theorem 4.4.9 limn→∞rn = 0. Whenever
Proof:If 0 < r < 1 if follows r−1 > 1. Why? Letting α =
Therefore, by the binomial theorem,
Therefore, limn→∞rn = 0 if 0 < r < 1. Now in general, if
0 by the first
An important theorem is the one which states that if a sequence converges, so
does every subsequence. You should review Definition 4.2.4 on Page 117 at this
Theorem 4.4.10 Let
be a sequence with
= x and let
be a subsequence. Then
Proof: Let ε > 0 be given. Then there exists nε such that if n > nε, then
Suppose k > nε.
Then nk ≥ k > nε
showing limk→∞xnk = x as claimed. ■
Theorem 4.4.11 Let
be a sequence of real numbers and suppose each
xn ≤ l
= x. Then x ≤ l
. More generally, suppose
are two sequences such that
= x and
= y. Then if xn ≤ yn
for all n sufficiently large, then x ≤ y.
Proof: Suppose not. Suppose that xn ≤ l but x > l. Then for n large enough,
a contradiction. The case where each xn ≥ l is similar. Consider now the last claim. For n
Since ε is arbitrary, it follows that y − x ≥ 0. ■