Let me see if I am following your logic.
Subtracting the 4th equation from the third:
S (6/7)=1+3/7+5/(7^2)+7/ (7^3)+....
S* 6/(7^2)=1/7+3/(7^2)+5/(7^3)+.....
---------------------------------------------
S*[ (6*7)/(7^2) ]=1+3/7+5/(7^2)+7/ (7^3)+....
S* 6/(7^2)=1/7+3/(7^2)+5/(7^3)+.....
----------------------------------------------------------
S*[ (42-6)/7^2 ] =1 + [3/7 - 1/7] + [ (5-2)/(7^2) ] + [ (7-5)/(7^3) ] + ...
-------------------------------------------------------------------
S (36/49)=1+2/7+2/(7^2)+....
_____________________________________________
OK, then what? Let's see ... Do you then factor out the 2? or are you factoring out 2/7?
S*(36/49) = 1 + (2/7)*[1 + 1/7 + 1/7^2 + 1/7^3 + ...]
ah, and that is your geometric progression with first term=1 and common ratio=1/7 ... where you use the formula 1/(1-r) ---> 1/(1-(1/7)) ---> 1/(6/7) ---> 7/6
So, S*(36/49) = 1 + (2/7)*(7/6)
S*(36/49) = 4/3
S = (4/3)*(49/36) = (1/3)*(49/9) = 49/27
Very clever. Kind of Gaussian. And no computers, you rebel! Sorry if I doubted you for a minute. I had to prove it to myself. Wow. That's very cool, Holden.
I like that. I was just shutting down the computer for the night, so I checked the board first. I'm pecking away at Infinite Jest again.
That is a very cool method you used, a real confidence builder where you attack it with brute force.
Your method reminds me a little of how one finds the sum of numbers between number x and number y, where you have
S = x + (x+1) + (x+2) + ... + (y-2) + (y-1) + y
Then you add the same sum (in reverse order):
S = x + (x+1) + (x+2) + ... + (y-2) + (y-1) + y
S = y + (y-1) + (y-2) + ... + (x+2) + (x+1) + x
--------------------------------------------------------
2S = (x+y) + (x+y) + (x+y) + ... + (x+y) + (x+y) + (x+y)
There are n terms, so 2S = n(x+y)
Then, dividing both sides by 2, S = [ n (x+y) ]/2
Thanks for the inspiration. It's good to be able to work such things out by hand without a full scale 8 gigabyte computer algebra system!
Post Scriptum:
If you ever want to check your results, and you can extract {a(n)} from the given terms of the series, go to
http://sagecell.sagemath.org/. For example, copy the following into the cell and press evaluate:
n = var('n')
h(n) = (n^2)*(1/7)^(n-1)
sum(h(n), n, 1, oo)
Note that oo represents "infinity", that is, two lowercase o.