2013-03-17, 07:43 PM
Figured it out. It wasn't partial sums.
-ln(1 - x) = sum from 0 to infinity of (x^(n+1))/(n + 1)
In my case, x = 1/3.
So, the sum is -ln(2/3).
-ln(1 - x) = sum from 0 to infinity of (x^(n+1))/(n + 1)
In my case, x = 1/3.
So, the sum is -ln(2/3).