James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

702 Chapter 11 Infinite Sequences and Series
monotonic sequence is convergent sa n − n l `d. But if a sequence is both bounded and
monotonic, then it must be convergent. This fact is proved as Theorem 12, but intuitively
you can understand why it is true by looking at Figure 12. If ha n j is increasing and
a n < M for all n, then the terms are forced to crowd together and approach some number
L.
M
a n
1 3
L
FIGURE 12
0 2
n
The proof of Theorem 12 is based on the Completeness Axiom for the set R of real
numbers, which says that if S is a nonempty set of real numbers that has an upper bound
M (x < M for all x in S), then S has a least upper bound b. (This means that b is an
upper bound for S, but if M is any other upper bound, then b < M.) The Completeness
Axiom is an expression of the fact that there is no gap or hole in the real number line.
12 Monotonic Sequence Theorem Every bounded, monotonic sequence is
convergent.
Proof Suppose ha n j is an increasing sequence. Since ha n j is bounded, the set
S − ha n | n > 1j has an upper bound. By the Completeness Axiom it has a least upper
bound L. Given « . 0, L 2 « is not an upper bound for S (since L is the least upper
bound). Therefore
a N . L 2 «
for some integer N
But the sequence is increasing so a n > a N for every n . N. Thus if n . N, we have
a n . L 2 «
so 0 < L 2 a n , «
since a n < L. Thus
| L 2 a n | , « whenever n . N
so lim n l ` a n − L.
A similar proof (using the greatest lower bound) works if ha n j is decreasing.
n
The proof of Theorem 12 shows that a sequence that is increasing and bounded above
is convergent. (Likewise, a decreasing sequence that is bounded below is convergent.)
This fact is used many times in dealing with infinite series.
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