In mathematics, the Mercator series or Newton–Mercator series is the Taylor series for the natural logarithm:

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

In summation notation,

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

The series converges to the natural logarithm (shifted by 1) whenever ${\displaystyle -1<x\leq 1}$ .

The series was discovered independently by Johannes Hudde^[1] and Isaac Newton. It was first published by Nicholas Mercator, in his 1668 treatise Logarithmotechnia.

The series can be obtained from Taylor's theorem, by inductively computing the n^th derivative of ${\displaystyle \ln(x)}$ at ${\displaystyle x=1}$ , starting with

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

Alternatively, one can start with the finite geometric series (${\displaystyle t\neq -1}$)

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

which gives

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

It follows that

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

and by termwise integration,

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

If ${\displaystyle -1<x\leq 1}$ , the remainder term tends to 0 as ${\displaystyle n\to \infty }$.

This expression may be integrated iteratively k more times to yield

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

where

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

and

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

are polynomials in x.^[2]

Setting ${\displaystyle x=1}$ in the Mercator series yields the alternating harmonic series

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

The complex power series

${\displaystyle \sum _{n=1}^{\infty }{\frac {z^{n}}{n}}=z+{\frac {z^{2}}{2}}+{\frac {z^{3}}{3}}+{\frac {z^{4}}{4}}+\cdots }$

is the Taylor series for ${\displaystyle -\log(1-z)}$ , where log denotes the principal branch of the complex logarithm. This series converges precisely for all complex number ${\displaystyle |z|\leq 1,z\neq 1}$. In fact, as seen by the ratio test, it has radius of convergence equal to 1, therefore converges absolutely on every disk B(0, r) with radius r < 1. Moreover, it converges uniformly on every nibbled disk ${\textstyle {\overline {B(0,1)}}\setminus B(1,\delta )}$, with δ > 0. This follows at once from the algebraic identity:

${\displaystyle (1-z)\sum _{n=1}^{m}{\frac {z^{n}}{n}}=z-\sum _{n=2}^{m}{\frac {z^{n}}{n(n-1)}}-{\frac {z^{m+1}}{m}},}$

observing that the right-hand side is uniformly convergent on the whole closed unit disk.

• John Craig