Question 982142
prove that. in a geometric series which has a sum to infinity. each term bears a
constant ratio to the sum of all the following terms
<pre>
We use {{{S[infinity]=a/(1-r)}}} where "a" = the first term and r = the
common difference:

Suppose the series is {{{a+ar+ar^2+ar^3+""*""*""*""+ar^(n+1)+ar^(n+2)+""*""*""*""}}}

The sum of all the terms following ar<sup>n+1</sup>, using the formula is

{{{ar^(n-2)/(1-r)}}}.

We want to show that the ratio of the nth term {{{ar^(n-1)}}} to the sum of all
the following terms {{{ar^(n-2)/(1-r)}}} is a constant:

That ratio is found by dividing:

  {{{ar^(n-1)}}}{{{"÷"}}}{{{ar^(n-2)/(1-r)}}} =

  {{{ar^(n-1)}}}{{{"×"}}}{{{(1-r)/(ar^(n-2))}}} =

That simplifies to 

 r(1-r) which is a constant.

So we have proved the proposition.

Edwin</pre>