The concept of the limit of a sequence was defined precisely by
Bolzano.^{1}
The following is the precise definition of what is meant by the limit of a sequence.
Definition 4.4.1A sequence
{a }
n
_{n=1}^{∞}converges to a,
lim a = a or a → a
n→ ∞ n n
if and only if for every ε > 0 there exists n_{ε}such that whenever n ≥ n_{ε},
|an − a| < ε.
Here a and a_{n}are assumed to be complex numbers but the same definition holds moregenerally.
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 a_{n} are complex numbers or later on vectors, the definition remains the
same. If a_{n} = x_{n} + iy_{n} and a = x + iy,
|an − a|
=
∘ -------2---------2
(xn − x) + (yn − y)
. Recall the way you
measure distance between two complex numbers.
Theorem 4.4.2If lim_{n→∞}a_{n} = a and lim_{n→∞}a_{n} = âthenâ = a.
Proof:Suppose â≠a. Then let 0 < ε <
|ˆa − a|
∕2 in the definition of the limit. It follows
there exists n_{ε} such that if n ≥ n_{ε}, then
In fact, this is true from the definition. Let ε > 0 be given. Let n_{ε}≥
√ ---
ε−1
. Then
if
√ ---
n > nε ≥ ε−1,
it follows that n^{2} + 1 > ε^{−1} and so
---1--
0 < n2 + 1 = an < ε..
Thus
|an − 0|
< ε whenever n 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 4.4.4Let a_{n} = n^{2}
Then in this case lim_{n→∞}a_{n} does not exist.
Example 4.4.5Let a_{n} =
(− 1)
^{n}.
In this case, lim_{n→∞}
(− 1)
^{n} does not exist. This follows from the definition. Let ε = 1∕2.
If there exists a limit, l, then eventually, for all n large enough,
Another very useful theorem for finding limits is the squeezing theorem.
Theorem 4.4.7Suppose lim_{n→∞}a_{n} = a = lim_{n→∞}b_{n}and a_{n}≤ c_{n}≤ b_{n}forall n large enough. Then lim_{n→∞}c_{n} = a.
Proof: Let ε > 0 be given and let n_{1} be large enough that if n ≥ n_{1},
|an − a| < ε∕2 and |bn − a| < ε∕2.
Then for such n,
|c − a| ≤ |a − a|+ |b − a| < ε.
n n n
The reason for this is that if c_{n}≥ a, then
|cn − a| = cn − a ≤ bn − a ≤ |an − a|+ |bn − a|
because b_{n}≥ c_{n}. On the other hand, if c_{n}≤ a, then
|cn − a| = a− cn ≤ a − an ≤ |a− an|+ |b− bn|. ■
As an example, consider the following.
Example 4.4.8Let
c ≡ (− 1)n 1
n n
and let b_{n} =
1n
, and a_{n} = −
1n
. Then you may easily show that
nl→im∞ an = nl→im∞ bn = 0.
Since a_{n}≤ c_{n}≤ b_{n}, it follows lim_{n→∞}c_{n} = 0 also.
Theorem 4.4.9 lim_{n→∞}r^{n} = 0. Whenever
|r|
< 1.
Proof:If 0 < r < 1 if follows r^{−1}> 1. Why? Letting α =
1
r
− 1, it follows
--1--
r = 1+ α .
Therefore, by the binomial theorem,
0 < rn =---1----≤ --1---.
(1+ α)n 1+ αn
Therefore, lim_{n→∞}r^{n} = 0 if 0 < r < 1. Now in general, if
|r|
< 1,
|rn|
=
|r|
^{n}→ 0 by the first
part. ■
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
point.
Theorem 4.4.10Let
{xn}
be asequence with lim_{n→∞}x_{n} = x and let
{xnk}
be a subsequence. Then lim_{k→∞}x_{nk} = x.
Proof: Let ε > 0 be given. Then there exists n_{ε} such that if n > n_{ε}, then
|xn − x|
< ε.
Suppose k > n_{ε}. Then n_{k}≥ k > n_{ε} and so
|xnk − x| < ε
showing lim_{k→∞}x_{nk} = x as claimed. ■
Theorem 4.4.11Let
{xn}
be a sequence ofreal numbers and suppose eachx_{n}≤ l
(≥ l)
and lim_{n→∞}x_{n} = x. Then x ≤ l
(≥ l)
. More generally, suppose
{xn}
and
{yn}
are two sequences such that lim_{n→∞}x_{n} = x and lim_{n→∞}y_{n} = y. Then if x_{n}≤ y_{n}for all n sufficiently large, then x ≤ y.
Proof: Suppose not. Suppose that x_{n}≤ l but x > l. Then for n large enough,
|xn − x| < x − l
and so
x− xn < x − l which implies xn > l
a contradiction. The case where each x_{n}≥ l is similar. Consider now the last claim. For n
large enough,
y − x ≥ (yn − ε)− (xn +ε) ≥ (yn − xn)− 2ε ≤ − 2ε
Since ε is arbitrary, it follows that y − x ≥ 0. ■