Posted by sarah on Friday, March 21, 2008 at 1:02am.
1) Log[1 + x] =
sum from n = 1 to infinity of
(-1)^(n+1) x^n/n
2) 1/[1+x] = sum from n=0 to infinity of x^n
x = 1/2, there is no n^(-2) in here.
3)
Differentiate both sides of the equation:
1/[1+x] = sum from n=0 to infinity of x^n
with respect to x.
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