Proving the Convergence of Continued Fractions

Date: 01/10/2001 at 09:27:29
From: Katy Fields
Subject: Continued Fractions

The Ask Dr. Math archive entry:

  22/7 as an Approximation for Pi 
  http://mathforum.org/dr.math/problems/faulkner4.1.98.html"    

assumes that the sequence of convergents 

  3 + 1/(7 + 1/(15 + 1/(1 + 1/(292 + 1/...)))) 

actually converges to pi.

How do you prove that the sequence actually converges? 

What I'm looking for is something similar to the proof in elementary 
calculus that the Taylor Series converges, a derivation of a formula 
that provides an upper bound on the difference between the convergent 
and the number toward which it converges.

If this question is beyond the scope of this website (I hope it 
isn't!), please tell me which books on this subject cover this and do 
so most clearly.

Thank you,
Katy

Date: 01/10/2001 at 17:06:24
From: Doctor Rob
Subject: Re: Continued Fractions

Thanks for writing to Ask Dr. Math, Katy.

The proof that the convergents to a simple continued fraction do 
converge to the value you started with goes like this.

Let x be irrational, and let its simple continued fraction have the 
form:

    a[0] + 1/(a[1] + 1/(a[2] + 1/(a[3] + 1/(a[4] + 1/(a[5] + ...)))))

Now define

     x[0]   = x
     a[0]   = greatest integer in x[0]
     p[-1]  = 1
     q[-1]  = 0
     p[0]   = a[0]
     q[0]    = 1
     x[n+1] = 1/(x[n]-a[n])
     a[n+1] = greatest integer in x[n+1]
     p[n+1] = a[n+1]*p[n] + p[n-1]
     q[n+1] = a[n+1]*q[n] + q[n-1]

These definitions imply that a[n] >= 1 for all n >= 1. Then the nth 
convergent to x is p[n]/q[n]. a[n] is called the nth partial quotient, 
and x[n] is called the nth complete quotient. Then the simple 
continued fraction of x is:

     a[0] + 1/(a[1] + 1/(a[2] + 1/(a[3] + 1/(a[4] + 1/(a[5] + ...)))))

p[n]/q[n] is the fraction you get by truncating this fraction at the 
point ... + 1/a[n])..., and simplifying. (This is the analogue of a 
partial sum of an infinite series.) You can prove by induction that 
for any value of n >= 0,

     p[n]*q[n-1] - p[n-1]*q[n] = (-1)^(n-1)

You can also prove by induction that for all n >= 2,

     q[n+1] >= q[n] + 1

This implies that as n -> infinity, q[n] -> infinity. You can also 
prove by induction that

     x = a[0] + 1/(a[1] + 1/(a[2] + ... + 1/(a[n-1] + 1/x[n])...))

Now you can prove that

     p[0]/q[0] < p[2]/q[2] < ... < x < ... < p[3]/q[3] < p[1]/q[1]

Thus the even convergents are increasing, but always less than x, and 
the odd convergents are decreasing, but always greater than x.

Now all you need to show is that the distance between consecutive 
convergents approaches zero, that is,

         lim    |p[n]/q[n]-p[n-1]/q[n-1]| = 0
     n->infinity

         lim    |p[n]*q[n-1]-p[n-1]*q[n]|/(q[n]*q[n-1]) = 0
     n->infinity

         lim    |(-1)^(n-1)|/(q[n]*q[n-1]) = 0
     n->infinity

         lim    1/(q[n]*q[n-1]) = 0
     n->infinity

This last is a corollary of the fact that q[n] -> infinity as 
n -> infinity.

That shows that the odd- and even-numbered convergents are approaching 
each other, with x caught in between, so both must approach x. Thus

       lim    p[n]/q[n] = x
   n->infinity

(This is the analogue of the limit of the partial sums of a convergent 
infinite series being set equal to the infinite sum.)

That is why it makes sense to write that an irrational number x equals 
the infinite simple continued fraction, so:

     x = a[0] + 1/(a[1] + 1/(a[2] + 1/(a[3] + 1/(a[4] + ...))))

By the way, if x is rational, the continued fraction will terminate.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   

Date: 01/16/2001 at 11:12:56
From: Katy Fields
Subject: Re: Continued Fractions

Hi Doctor Rob.

I have absorbed your excellent and much appreciated proof that 
convergents converge.

I respectfully submit that you might have overlooked a minor point 
(see below) that has no effect on the conclusion.

Thank you again for your very generous help.

Sincerely,
Katy Fields


***** (Part I: Pointing out the applicable part of the proof)

The proof depends upon

     q[n+1] >= q[n] + 1

which follows from

     q[n+1] = a[n+1]*q[n] + q[n-1]

and

     a[n+1] > 1 and q[n-1] > 1

where a[n+1] > 1 follows from the recursion relation that defines 
x[n+1] as

     x[n+1] = 1/(x[n]-a[n])


***** (Part II: Criticism of above)

Nothing precludes an arbitrary continued fraction from having one or 
more terms equal to zero.  In other words, we cannot ignore a[n+1] = 
0. To illustrate this point, I can generate such a continued fraction 
from any other continued fraction by simply inserting a zero term 
anywhere I like.


***** (Part III: Resolution)

This exception can be addressed using the following simple 
"contraction" theorem, which allows us to replace any simple continued 
fraction having one or more terms equal to zero by an equal simple 
continued fraction whose terms are all non-zero.

A simple continued fraction with n terms, whose kth term is zero, is 
equal to a simple continued fraction of two less terms (ie. n-2 terms) 
that differs from the original simple continued fraction by replacing 
three terms (the k-1st, kth, and k+1st terms) by one term equal to 
their sum (a[k-1] + 0 + a[k+1]).


***** (Part IV: Corollary)

The above contraction theorem has the following simple corollary:

A simple continued fraction that has two consecutive terms that are 
zero is equal to a simple continued fraction of two less terms that 
differs from the original simple continued fraction by simply the 
omission of the two zero terms.


***** (Part V: Questions)

If A[x] is the continued fraction that is generated from the real 
number x by the usual recursive formula, what is x' = A[x, n, 0], the 
value of the real number that results by inserting a zero before the 
nth term of A[x]?

More generally, what is x' = A[x, n, z], the value of the real number 
that results by inserting the integer value z before the nth term of 
A[x]?

Similarly, what is x' = A[x, delete n], the value of the real number 
that results by deleting  the nth term of A[x]?

And similarly, what is x' = A[x, n, substitute z], the value of the 
real number that results by substituting (vs inserting) the integer 
value z for the nth term of A[x]?

I appreciate that these are just a few of the questions that 
mathematicians have addressed. Rather than try to reinvent all this 
mathematics and stumble upon all the pitfalls along the way, I'd like 
to find a clear exposition of it. In particular, my interest in this 
field has been sparked by Calvin Clawson's book _Mathematical 
Mysteries_, and I would like to find a book that would instruct me on 
not just simple continued fractions but also general continued 
fractions, continued radicals, etc.

One particular curiosity I'd like to address comes from the 
realization that the general continued fraction generalizes the simple 
continued fraction - which amounts to Euclid's GCD algorithm when the 
real number we start with is rational. Since Euclid's GCD algorithm is 
the basis for most of the elementary number theory I know, I assume 
that the general continued fraction is the basis for much more. I'd 
like to learn more about how the theory of general continued fractions 
has been developed.

Date: 01/16/2001 at 12:41:36
From: Doctor Rob
Subject: Re: Continued Fractions

Thanks for writing back.

Continued fractions with 0 partial quotients after the first one 
cannot occur by the process described. If

     a[n+1] = 0 = Floor(x[n+1])

then

     0 <=    x[n+1]     < 1
     0 <= 1/(x[n]-a[n]) < 1

                   1 < x[n] - a[n]
                a[n] < x[n] - 1
         Floor(x[n]) < x[n] - 1
     Floor(x[n]) + 1 < x[n]

which is impossible by the definition of the Floor (or greatest 
integer) function. It is possible for a[0] <= 0, but for larger 
subscripts for a[n], all these values must be 1 or larger.

This means that the case that was troubling you, but which you 
resolved, is not an issue.

The only book that I know of that contains the information you seek 
is written in German. It is _Die Lehre von Kettenbruche_, by Oskar 
Perron. It seems likely that there are books in English covering this, 
but I don't know about them.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/