Infinite Geometric Progression

Date: 10/02/97 at 20:56:26
From: Franklin Lum
Subject: Repetend or infinite repeating decimal

Hi Dr. Math,

I teach eightt grade math. I can convert infinite repeating numbers 
from fractions to decimals with no trouble, e.g.  3/11  = .2727 or 
.27 bar on the top 27.  My problem is how do you go about solving the 
.27 bar on both numbers into fractions?  How do you do that?  Could 
you tell me the steps?  
  
Please help.  

Judiana

Date: 10/03/97 at 08:42:59
From: Doctor Jerry
Subject: Re: Repetend or infinite repeating decimal

The idea is to use the formula for the sum of an infinite geometric 
progression. Let me first remind you of that.

A geometric progression is one that generates each new term by 
multiplying by a fixed ratio, usually named r. If the first term 
is a, then the second term is a*r, the third term is a*r*r = a*r^2, 
and so on.

Let S be the sum a+a*r+a*r^2+a*r^3+...

Unless a happens to be zero, the sum is infinite or not well defined 
if the absolute value of r is 1 or bigger:

   a = 1, r = 2:  1+1*2+1*2^2+1*2^3+... = 1+2+4+8+...
   a = 1, r = -2: 1-1*2+2*2^2-1*2^3+... = 1-2+4-8+...

Suppose a is not zero and |r|<1 (this is the case in the repeating 
decimals).

Okay, the sum a+a*r+a*r^2+a*r^3+... is given by

     S = a/(1-r).

This is easily seen.  Just write

     S = a+a*r+a*r^2+a*r^3+...

Now multiply both sides by r

   r*S = a*r+a*r^2+a*r^3+...

Now substract,

 S-r*S = a.

So,  S = a/(1-r).

Now for 0.272727...  This (and other such repeating decimals) can be 
written as 

   27/100+(27/100)*(1/100)+(27/100)(1/100)^2+...

So,

   0.272727... = a/(1-r) = (27/100)/(1-1/100) = 27/99 = 3/11.

Neat, huh?  I'm not patting myself on the back. The above is just my 
memory of some earlier person's work. I enjoy doing the demo, though.

-Doctor Jerry,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/