3.1. Sequences
Proposition 3.1.4: Convergent Sequences are Bounded
Proof:
Let's prove uniqueness first. Suppose the sequence has two limits, a and a'. Take any
> 0. Then there is an
integer N such that:
| aj - a | <if j > N. Also, there is another integer N' such that![]()
| aj - a' | <if j > N'. Then, by the triangle inequality:![]()
| a - a' | = | a - aj + aj - a' |if j > max{N,N'}. Hence | a - a' | < 2
|aj - a | + | aj - a' |
<+
= 2
![]()
for any
> 0. But that implies
that a = a', so that the limit is indeed unique.
Next, we prove boundedness. Since the sequence converges, we can take, for
example,
= 1. Then
| aj - a | < 1if j > N. Fix that number N. We have that
| aj |for all j > N. Define| aj - a | + | a | < 1 + |a|
M = max{|a1|, |a2|, ...., |aN|, (1 + |a|)}Then | aj | < M for all j, i.e. the sequence is bounded as required.

Interactive Real Analysis
be a
convergent sequence. Then the sequence is bounded, and the limit is
unique.
|aj - a | + | aj - a' |