Feb 18, 2014 Thread starter Admin #1 anemone MHB POTW Director Staff member Feb 14, 2012 3,963 Let $m$ and $n$ be real numbers with $\dfrac{1}{m}-\dfrac{1}{n}=1$, $0<m\le \dfrac{1}{2}$. Show that $m+\dfrac{m^2}{2}+\dfrac{m^3}{3}+\cdots=n-\dfrac{n^2}{2}+\dfrac{n^3}{3}-\cdots$.
Let $m$ and $n$ be real numbers with $\dfrac{1}{m}-\dfrac{1}{n}=1$, $0<m\le \dfrac{1}{2}$. Show that $m+\dfrac{m^2}{2}+\dfrac{m^3}{3}+\cdots=n-\dfrac{n^2}{2}+\dfrac{n^3}{3}-\cdots$.
Feb 18, 2014 #2 Random Variable Well-known member MHB Math Helper Jan 31, 2012 253 Spoiler $ - \log(1-m) = m+\dfrac{m^2}{2}+\dfrac{m^3}{3}+\cdots$ $(-1 \le m <1)$ $ \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) $
Spoiler $ - \log(1-m) = m+\dfrac{m^2}{2}+\dfrac{m^3}{3}+\cdots$ $(-1 \le m <1)$ $ \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) $
Feb 18, 2014 Thread starter Admin #3 anemone MHB POTW Director Staff member Feb 14, 2012 3,963 Thank you for participating, Random Variable! Your proof is correct, well done!