A little later, the limit of functions will be important. A sequence is just a special kind of
function and it turns out that it is easier to consider the limit of a sequence. This also
helps considerably in understanding certain other concepts like continuity of a
function also presented later. This is why I am including this topic early in the
book.
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 3.2.1A sequence
{a }
n
_{n=1}^{∞}converges to a, written
lim an = a or an → a
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 real 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.
Theorem 3.2.2If 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
0 < --1---= a < ε..
n2 + 1 n
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 3.2.4Let a_{n} = n^{2}
Then in this case lim_{n→∞}a_{n} does not exist.
Example 3.2.5Let 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 3.2.7Suppose lim_{n→∞}a_{n} = a = lim_{n→∞}b_{n}and a_{n}≤ c_{n}≤ b_{n}for all 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,
|cn − a| ≤ |an − a|+ |bn − a| < ε.
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 3.2.8Let
cn ≡ (− 1)n 1
n
and let b_{n} =
1n
, and a_{n} = −
1n
. Then you may easily show that
lnim→∞ an = lnim→∞ bn = 0.
Since a_{n}≤ c_{n}≤ b_{n}, it follows lim_{n→∞}c_{n} = 0 also.
Theorem 3.2.9 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
r =--1--.
1 + α
Therefore, by the binomial theorem,
0 < rn =---1--n-≤ --1---.
(1+ α) 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 3.0.4 on Page 208 at this
point.
Theorem 3.2.10Let
{x }
n
be asequence with lim_{n→∞}x_{n} = x and let
{x }
nk
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
|x − x|
n
< 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 3.2.11Let
{x }
n
be a sequence ofreal numbers and suppose eachx_{n}≤ l
(≥ l)
and lim_{n→∞}x_{n} = x. Then x ≤ l
(≥ l)
. More generally, suppose
{x }
n
and
{y }
n
are two sequences such that lim_{n→∞}x_{n} = x and lim_{n→∞}y_{n} = y. Then ifx_{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. ■
Sometimes the limit of a sequence does not exist. However, there are two
things which always exist for any sequence. Suppose
{an}
is a sequence, and let
A_{n}≡ inf
{ak : k ≥ n}
. Then
{An }
is an increasing sequence in the sense A_{n}≤ A_{n+1}
because the sets
{ak : k ≥ n}
are getting smaller as n increases. Thus either
{An}
is
bounded above and lim_{n→∞}A_{n} is a real number equal to sup_{n}
{An }
or they are
not bounded above and in this case, we say liminf _{n→∞}A_{n} = ∞. Similarly, if
B_{n}≡ sup
{ak : k ≥ n}
, the B_{n} are decreasing and we can also consider lim_{n→∞}B_{n}
which equals −∞ if not bounded below and some real number equal to inf _{n}
{Bn }
otherwise. This explains the following definition.
Definition 3.2.12Let A_{n},B_{n}be as just described relative to a sequence
{an}
.Then
lim nin→f∞ an ≡ nl→im∞ An ≡ lnim→∞ (inf{ak : k ≥ n })
lim sup an ≡ lim Bn ≡ lim (sup {ak : k ≥ n})
n→∞ n→ ∞ n→∞
Here is a useful proposition.
Proposition 3.2.13Let lim_{n→∞}a_{n} = a > 0 and suppose that each b_{n}> 0.Then
lim sup anbn = alim sup bn.
n→∞ n→∞
Proof: This follows from the definition. Let λ_{n} = sup
{akbk : k ≥ n}
. For all
n large enough, a_{n}∈
(a− ε,a+ ε)
where ε is small enough that a − ε > 0.
Therefore,
λn ≥ sup{bk : k ≥ n}(a− ε)
for all n large enough. Then
lim sup anbn = lim λn ≡ lim sup anbn
n→ ∞ n→∞ n→∞
≥ lni→m∞ (sup{bk : k ≥ n}(a− ε))
= (a− ε)lim sup b
n→∞ n
Similar reasoning shows
lim sup anbn ≤ (a + ε)lim sup bn
n→∞ n→ ∞
Now since ε > 0 is arbitrary, the conclusion follows. ■