# 墨卡托級數

${\displaystyle \ln(1+x)\;=\;x\,-\,{\frac {x^{2}}{2}}\,+\,{\frac {x^{3}}{3}}\,-\,{\frac {x^{4}}{4}}\,+\,\cdots .}$

${\displaystyle \ln(1+x)\;=\;\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}x^{n}.}$

## 推導

${\displaystyle {\frac {d}{dx}}\ln x={\frac {1}{x}}.}$

${\displaystyle 1-t+t^{2}-\cdots +(-t)^{n-1}={\frac {1-(-t)^{n}}{1+t}}}$

${\displaystyle {\frac {1}{1+t}}=1-t+t^{2}-\cdots +(-t)^{n-1}+{\frac {(-t)^{n}}{1+t}}.}$

${\displaystyle \int _{0}^{x}{\frac {dt}{1+t}}=\int _{0}^{x}\left(1-t+t^{2}-\cdots +(-t)^{n-1}+{\frac {(-t)^{n}}{1+t}}\right)\,dt}$

${\displaystyle \ln(1+x)=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-\cdots +(-1)^{n-1}{\frac {x^{n}}{n}}+(-1)^{n}\int _{0}^{x}{\frac {t^{n}}{1+t}}\,dt.}$

${\displaystyle -xA_{k}(x)+B_{k}(x)\ln(1+x)=\sum _{n=1}^{\infty }(-1)^{n-1}{\frac {x^{n+k}}{n(n+1)\cdots (n+k)}},}$

${\displaystyle A_{k}(x)={\frac {1}{k!}}\sum _{m=0}^{k}{k \choose m}x^{m}\sum _{l=1}^{k-m}{\frac {(-x)^{l-1}}{l}}}$

${\displaystyle B_{k}(x)={\frac {1}{k!}}(1+x)^{k}}$

## 特例

${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k}}=\ln 2.}$

## 複數級數

${\displaystyle z\,-\,{\frac {z^{2}}{2}}\,+\,{\frac {z^{3}}{3}}\,-\,{\frac {z^{4}}{4}}\,+\,\cdots }$

## 參考資料

1. ^ Medina, Luis A.; Moll, Victor H.; Rowland, Eric S. Iterated primitives of logarithmic powers. 2009. .