# Mercator series

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

$\ln(1+x)=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+\cdots$ $\ln(1+x)=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}x^{n}.$ The series converges to the natural logarithm (shifted by 1) whenever $-1 .

## History

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

## Derivation

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

${\frac {d}{dx}}\ln(x)={\frac {1}{x}}.$ Alternatively, one can start with the finite geometric series ($t\neq -1$ )

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

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

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

$\ln(1+x)=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-\cdots +(-1)^{n-1}{\frac {x^{n}}{n}}+(-1)^{n}\int _{0}^{x}{\frac {t^{n}}{1+t}}\ dt.$ If $-1 , the remainder term tends to 0 as $n\to \infty$ .

This expression may be integrated iteratively k more times to yield

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

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

$B_{k}(x)={\frac {1}{k!}}(1+x)^{k}$ are polynomials in x.

## Special cases

Setting $x=1$ in the Mercator series yields the alternating harmonic series

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

$\sum _{n=1}^{\infty }{\frac {z^{n}}{n}}=z+{\frac {z^{2}}{2}}+{\frac {z^{3}}{3}}+{\frac {z^{4}}{4}}+\cdots$ is the Taylor series for $-\log(1-z)$ , where log denotes the principal branch of the complex logarithm. This series converges precisely for all complex number $|z|\leq 1,z\neq 1$ . 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:

$(1-z)\sum _{n=1}^{m}{\frac {z^{n}}{n}}=z-\sum _{n=2}^{m}{\frac {z^{n}}{n(n-1)}}-{\frac {z^{m+1}}{m}},$ observing that the right-hand side is uniformly convergent on the whole closed unit disk.