*Proof*.
We use induction. The statements are both true when

.
If the second statement is true for

, then

Similarly, the first statement is true for

if
it is true for

.

*Proof*.
For any

, the number

is a partial convergent of

. By Proposition

5.1.13 the
even convergents

form a strictly

*increasing* sequence
and the odd convergents

form a strictly

*decreasing*
sequence. Moreover, the even convergents are all

and the
odd convergents are all

. Hence

and

both
exist and

. Finally, by
Proposition

5.1.7

so

.

We define

*Example 5.2*
We illustrate the theorem with

.
As in the proof of Theorem

5.2.6,
let

be the

th partial convergent
to

. The

with

odd converge down to

whereas the

with

even converge up to

**Theorem 5.2**
*
Let
be a sequence of real numbers
such that
for all
, and for each
, set
Then
exists if and only if the sum
diverges.*
*Proof*.
We only prove that if

diverges then

exists. A proof of the converse can be found in
[#!wall!#, Ch. 2, Thm. 6.1].

Let
be the sequence of ``denominators''
of the partial convergents, as defined in Section 5.1.1,
so
,
, and for
,

As we saw in the proof of Theorem

5.2.6,
the limit

exists provided that
the sequence

diverges to positive infinity.

For
even,

and for

odd,

Since

for

, the sequence

is increasing, so

for all

.
Applying this fact to the above expressions for

, we see that
for

even

and for

odd

If
diverges, then at least one of
or
must diverge. The
above inequalities then imply that
at least one of the sequences
or
diverge to infinity.
Since
is an increasing sequence, it follows
that
diverges to infinity.

*Example 5.2*
Let

for

and

. By the
integral test,

diverges, so by Theorem

5.2.8
the continued fraction

converges.
This convergence is very slow, since, e.g.

yet

**Theorem 5.2**
*
Let
be a real number. Then
is the value
of the (possibly infinite) simple continued fraction
produced by the continued fraction procedure.*
*Proof*.
If the sequence is finite then some

and the
result follows by Lemma

5.2.5.
Suppose the sequence is infinite.
By Lemma

5.2.5,

By Proposition

5.1.5 (which we apply in a case when
the partial quotients of the continued fraction are not integers!),
we have

Thus if

, then

Thus

In the inequality we use that

is the integer part of

, and is hence

, since

.

This corollary follows from the proof of the above theorem.

**Corollary 5.2** (Convergence of continued fraction)

*Let
define a simple continued
fraction, and let
be its value.
Then for all
,
*

**Proposition 5.2**
*If
is a rational number then the sequence
produced by the continued fraction procedure terminates.*
*Proof*.
Let

be the continued fraction representation
of

that we obtain using Algorithm

1.1.13, so the

are the partial quotients at each step.
If

, then

is an integer, so we may assume

.
Then

If

then

and

,
which will not happen using Algorithm

1.1.13, since
it would give

for the continued fraction of
the integer

.
Thus

, so in the continued fraction
algorithm we choose

and

.
Repeating this argument enough times proves the claim.

William
2007-06-01