Monotone sequences

The simplest type of a sequence is a monotone sequence. Checking that a monotone sequence
converges is as easy as checking that it is bounded. It is also easy to find the limit for a convergent
monotone sequence, provided we can find the supremum or infimum of a countable set of numbers.
Definition 2.1.9. A sequence fxng is monotone increasing if xn  xn+1 for all n 2 N. A sequence
fxng is monotone decreasing if xn  xn+1 for all n 2 N. If a sequence is either monotone increasing
or monotone decreasing, we simply say the sequence is monotone. Some authors also use the word
monotonic.
Theorem 2.1.10. A monotone sequence fxng is bounded if and only if it is convergent.
Furthermore, if fxng is monotone increasing and bounded, then
lim
n!¥
xn = supfxn j n 2 Ng:
If fxng is monotone decreasing and bounded, then
lim
n!¥
xn = inffxn j n 2 Ng:
Proof. Let us suppose that the sequence is monotone increasing. Suppose that the sequence is
bounded. That means that there exists a B such that xn  B for all n, that is the set fxn j n 2 Ng is
bounded. Let
x := supfxn j n 2 Ng:
Let e > 0 be arbitrary. As x is the supremum, then there must be at least one n0 2 N such that
xn0 > x􀀀e (because x is the supremum). As fxng is monotone increasing, then it is easy to see (by
induction) that xn  xn0 for all n  n0. Hence
jxn􀀀xj = x􀀀xn  x􀀀xn0 < e:
2.1. SEQUENCES AND LIMITS 43
Hence the sequence converges to x. We already know that a convergent sequence is bounded, which
completes the other direction of the implication.
The proof for monotone decreasing sequences is left as an exercise.
Example 2.1.11: Take the sequence fp1
ng.
First we note that p1
n > 0 and hence the sequence is bounded from below. Let us show that it
is monotone decreasing. We start with
p
n+1 
p
n (why is that true?). From this inequality we
obtain.
1
p
n+1

1
p
n
:
So the sequence is monotone decreasing, bounded from below (and hence bounded). We can apply
the theorem to note that the sequence is convergent and that in fact
lim
n!¥
1
p
n
= inf

1
p
n

:
We already know that the infimum is greater than or equal to 0, as 0 is a lower bound. Suppose we
take a number b  0 such that b  p1
n for all n. We can square both sides to obtain
b2 
1
n
;
for all n 2 N. We have seen before that this implies that b2  0 (a consequence of the Archimedean
property). As we also have b2  0, then b2 = 0 and hence b = 0. Hence b = 0 is the greatest lower
bound and hence the limit.
Example 2.1.12: Be careful however. You have to show that a monotone sequence is bounded
in order to use Theorem 2.1.10. For example, take the sequence f1+1=2+


+1=ng. This is a
monotone increasing sequence that grows very slowly. We will see, once we get to series, that this
sequence has no upper bound and so does not converge. It is not at all obvious that this sequence
has no bound.
A common example of where monotone sequences arise is the following proposition. The proof
is left as an exercise.
Proposition 2.1.13. Let S  R be a nonempty bounded set. Then there exist monotone sequences
fxng and fyng such that xn;yn 2 S and
sup S = lim
n!¥
xn and inf S = lim
n!¥
yn: