If is an infinite continued
fraction, I want to define its *value* to be the limit of the
convergents:

For this to make sense, I need to show that this limit exists.

In what follows, take as given an infinite continued fraction . Let

* Proposition.*

(a) .

(b) .

* Proof.* (a)

(b)

* Proposition.*

That is, the odd convergents get smaller, the even convergents get bigger, and any odd convergent is bigger than any even convergent.

* Proof.* If k is even, then

This shows that the even terms get bigger.

If k is odd, then

This shows that the odd terms get smaller.

Finally, I have to show any odd term is bigger than any even term. Note that it's true if the terms are adjacent:

Next, do the general case. Let be an odd term and let be an even term. Then

The first inequality is true because odd terms decrease. The second
inequality is true because I just observed that an odd term is bigger
than an *adjacent* even. Finally, the last inequality is true
because even terms increase.

I've shown that any odd term is bigger than any even term, and that completes the proof.

Here's a numerical example. The first ten convergents for the golden ratio are

The odd convergents get smaller, the even convergents get bigger, and any odd convergent is bigger than any even convergent.

* Proposition.* for all .

* Proof.* I'll induct on k. , so the result holds for . Take , and assume it holds for numbers less than
or equal to k. I'll prove that it holds for .

(The last inequality used .) This completes the induction step.

* Theorem.* Let be an infinite continued
fraction with for , and let be the k-th convergent. Then

* Proof.* Consider the sequence of odd
convergents

This is a decreasing sequence of numbers, and it's bounded below --- by any even convergent, for example. A standard result from analysis (see, for example, Theorem 3.14 of [1]) asserts that such a sequence must have a limit, so

Likewise, consider the sequence of even convergents:

This is an increasing sequence of numbers that's bounded above --- by any odd convergent, for example. The result from analysis mentioned above says that the sequence has a limit:

I have to show that the two limits agree.

The previous result implies that and , so

Now let . , so by the Squeezing Theorem of calculus,

Since the odd and even terms approach the same limit, exists.

* Definition.* .

What can I say about the value of an infinite continued fraction?

* Theorem.* Let be an infinite continued
fraction with for . Then is irrational.

* Proof.* Write for short. I want to
show that x is irrational. Suppose on the contrary that , where p and q are integers. I
will show this leads to a contradiction.

Since the odd convergents are bigger than x and the even convergents are smaller than x,

Then

Notice that this inequality is true for all k, and that is an *integer*. But q
is fixed, and ,
so if I make k sufficiently large eventually will become bigger than q. Then will be a fraction less than
1, and I have an *integer* caught between 0 and a
fraction less than 1. Since this is impossible, x can't be
rational.

Now I know that every infinite continued fraction made of positive integers represents an irrational number. The converse is also true, and it gives an algorithm for computing the continued fraction expansion.

* Proposition* Let be irrational. Let , and

Then for all ,

Thus, represents the "infinite tail" of the continued fraction.

* Proof.* For , the claim is that , which is true by definition.

Assume that the result holds for n. Then

Then

This proves the result for , so the result is true for by induction.

* Theorem.* Let be irrational. Let , and

Then

* Proof.*

* Step 1.* is irrational for .

Since x is irrational and , the result is true for .

Assume that and that the result is true for . I want to show that is irrational.

Suppose on the contrary that , where . Then

Now all the 's are clearly integers (since means they're outputs of the greatest integer function), so is the sum of an integer and a rational number. Therefore, it's rational, so is rational, contrary to the induction hypothesis.

It follows that is irrational. By induction, is irrational for all .

* Step 2.* The 's are positive integers for .

I already observed that the 's are integers.

Let . Since , the definition of the greatest integer function gives

But is irrational, so . Hence,

Since , this proves that the 's are positive integers for .

* Step 3.*

First, I'll get a formula for x in terms of the p's, q's, and a's.

Then I'll find and show that it's less than something which goes to 0.

Recall the recursion formulas for convergents:

The right sides only involve terms up to (and p's and q's of smaller indices). Therefore, the following fractions have the same p's and q's through index k:

Using the preceding proposition and the recursion formula for convergents, I get

Therefore,

Take absolute values:

Now

Therefore,

By an earlier lemma, and , so

Now , so by the Squeezing Theorem

This implies that

* Example.* Compute the first 5 terms of the
continued fraction expansion of .

Here are the first two steps:

Continuing in this way, I obtain:

* Example.* Compute the first 5 terms of the
continued fraction expansion of .

Here are the first 5 terms:

In fact, the continued fraction expansion for is .

* Theorem.* The continued fraction expansion of
an irrational number is unique.

* Proof.* Suppose there are two continued
fractions for the irrational number x, where and for :

I want to show that for all k.

Recall that the even convergents are smaller than x and the odd convergents are greater than x.

Therefore,

Now

Thus, is an integer less than x, and the next larger integer is greater than x. This means that .

The same reasoning applies to the b's. Therefore, , so .

Hence,

I can continue in the same way to show that for all k.

Here's a summary of some of the important results on infinite continued fractions:

1. An irrational number has a unique infinite continued fraction expansion.

2. The algorithm for computing the continued fraction expansion of an irrational number x is:

Then

3. If is the continued fraction expansion of an irrational number, then is an integer, and is a positive integer for .

4. If is a continued fraction expansion and for , then is irrational.

5. If is the continued fraction expansion of an irrational number and and are defined by the recursion formulas for convergents, then

* Example.* Compute the first 6 convergents in
the continued fraction expansion for .

Use them to illustrate the last result for .

Now

Thus, in this case,

[1] Walter Rudin, *Principles of Mathematical Analysis* ( edition). New York: McGraw-Hill Book
Company, 1976.

Copyright 2019 by Bruce Ikenaga