3.1 Sequences and Series 1113.1.4 More About Limits of Sequences
We continue our discussion about limits of sequences with three more topics: the method
of passing to the limit in a recurrence relation, the Cesàro–Stolz theorem, and Cantor’s
nested intervals theorem. We illustrate the first with the continued fraction expansion of
the golden ratio.
Example.Prove that
1 +√
5
2
= 1 +
1
1 +
1
1 +
1
1 +
1
1 +···
.
Solution.A close look at the right-hand side shows that it is the limit of a sequence(xn)n
subject to the recurrence relationx 1 =1,xn+ 1 = 1 +x^1 n. If this sequence has a finite limit
L, then passing to the limit on both sides of the recurrence relation yieldsL= 1 +L^1.
BecauseLcan only be positive, it must be equal to the golden ratio.
But does the limit exist? Investigating the first terms of the sequence we see that
x 1 <x 3 <1 +
√
5
2
<x 4 <x 2 ,and we expect the general situation to be
x 1 <x 3 <···<x 2 n+ 1 <···<1 +
√
5
2
<···<x 2 n<x 2 n− 2 <···<x 2.This can be proved by induction. Firstly, ifx 2 n+ 1 <^1 +
√ 5
2 , thenx 2 n+ 2 = 1 +1
x 2 n+ 1> 1 +
2
1 +
√
5
= 1 +
√
5 − 1
2
=
1 +
√
5
2
,
and by a similar computation, ifx 2 n+ 2 >^1 +
√ 5
2 , thenx^2 n+^3 <1 +√ 5
2. Secondly,xn+ 2 = 2 −1
xn+ 1,
and the inequalityxn+ 2 >xnis equivalent tox^2 n−xn− 1 <0, which holds if and
only ifxn<^1 +
√
5
2. Now an inductive argument shows that(x^2 n+^1 )nis increasing and
(x 2 n+ 2 )nis decreasing. Being bounded, both sequences are convergent. Their limits are
positive, and both should satisfy the equationL= 2 −L^1 + 1. The unique positive solution
to this equation is the golden ratio, which is therefore the limit of both sequences, and
consequently the limit of the sequence(xn)n.