Recent content by zodiak770

Brilliant--thank you guys so much. This problem had been gnawing at me for almost a week. In case any of you were wondering, this wasn't a homework problem--it was motivated by a google interview question. Check it out...

well, my first thought is express 1 - log(2) = log(e/2)=log( 1+(e-2)/2 ), ie, letting u be (e-2)/2, and then we end up with the = sum { (-1)^(n+1)*((e-2)/2)^n/n } from n=0 to infinity... is this the right track?

yes--that's what I meant. Sorry for not making it more readable.

sorry, what n(n+1) in the denominator? Also, there was a typo in my original post which I have since corrected--the first sum had terms of the form 2^(n+1)*n/(n+1), and it should have been 1/2^(n+1)*n/(n+1).

Prove that sum 1/2^(n+1)*n/(n+1), from n=0 to infinity, converges to 1 - log(2), where log stands for the natural logarithm. I know that the Taylor series for log(x) about x=1 is sum (-1)^(n+1)*(x-1)^n/n, but I don't see how these two statements are consistent. Thanks for any pointers!