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- Feb 14, 2012

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- Feb 14, 2012

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- Jan 31, 2012

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$ \log(1+n) = n-\dfrac{n^2}{2}+\dfrac{n^3}{3}-\cdots $ $( -1 < n \le 1)$

$ \displaystyle \frac{1}{m}- \frac{1}{n} = 1 \implies n = \frac{m}{1-m} $

Because of the restriction on $m$, the values of $n$ fall between $0$ and $1$ including $1$.

And we have

$ \displaystyle \log(1+n) = \log \left( 1 + \frac{m}{1-m} \right) = \log \left(\frac{1}{1-m} \right) = - \log(1-m) $

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Thank you for participating, **Random Variable**! Your proof is correct, well done!