Question 982376
<pre>
Instead of doing your problem for you, I will do one EXACTLY IN EVERY
DETAIL like yours, so that when you do yours, you can use this as a
model.

I will find the sum of this infinite geometric series instead:

{{{2/3-4/9+8/27-16/81+""-""*""*""*""}}}

We should first check to see if it really is a geometric series, for sometimes
books contain errors/typos.  If it is a geometric series then this must be true:

{{{matrix(1,2,2nd,term)/matrix(1,2,1st,term)}}}{{{""=""}}}{{{matrix(1,2,3rd,term)/matrix(1,2,2nd,term)}}}{{{""=""}}}{{{matrix(1,2,4th,term)/matrix(1,2,3rd,term)}}}

So we calculate the three ratio to see if they are all equal:

{{{matrix(1,2,2nd,term)/matrix(1,2,1st,term)}}}{{{""=""}}}{{{(-4/9)/(2/3)}}}{{{""=""}}}{{{expr(-4/9)*expr(3/2)}}}{{{""=""}}}{{{-12/18}}}{{{""=""}}}{{{-2/3}}}

{{{matrix(1,2,3rd,term)/matrix(1,2,2nd,term)}}}{{{""=""}}}{{{(8/27)/(-4/9)}}}{{{""=""}}}{{{expr(8/27)*expr(-9/4)}}}{{{""=""}}}{{{-72/108}}}{{{""=""}}}{{{-2/3}}}

{{{matrix(1,2,4th,term)/matrix(1,2,3rd,term)}}}{{{""=""}}}{{{(-16/81)/(8/27)}}}{{{""=""}}}{{{expr(-16/81)*expr(27/16)}}}{{{""=""}}}{{{-432/1296}}}{{{""=""}}}{{{-2/3}}}

They are all the same so it is a geometric series with common ratio = {{{-2/3}}}

It is a geometric series with a defined sum when taken to infinitely many
terms because the ratio {{{r=-2/3}}} is greater than -1 and less than +1.

{{{S[infinity]}}}{{{""=""}}}{{{a[1]/(1-r)}}}, where a<sub>1</sub> = 1st term = {{{2/3}}}

{{{S[infinity]}}}{{{""=""}}}{{{(2/3)/(1-(-2/3))}}}{{{""=""}}}{{{(2/3)/(1+2/3)}}}{{{""=""}}}{{{(2/3)/(3/3+2/3)}}}{{{""=""}}}{{{(2/3)/(5/3)}}}{{{""=""}}}{{{expr(2/3)expr(3/5)}}}{{{""=""}}}{{{6/15=2/5}}}

Answer to this problem = {{{2/5}}}

Now do your problem EXACTLY the same way.

Edwin</pre>