Prove that m+m²/2+...=n-n²/2+...

[anemone]

Let $m$ and $n$ be real numbers with $\dfrac{1}{m}-\dfrac{1}{n}=1$, $0<m\le \dfrac{1}{2}$. Show that $m+\dfrac{m^2}{2}+\dfrac{m^3}{3}+\cdots=n-\dfrac{n^2}{2}+\dfrac{n^3}{3}-\cdots$.

 

[Random Variable]

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) $

 

[anemone]

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