## Convergence of Infinite Continued Fractions

Lemma 5.2   For every such that is defined, we have

and if then

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 .

Theorem 5.2 (Continued Fraction Limit)   Let be a sequence of integers such that for all , and for each , set Then exists.

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