logarithm series

The derivative of  $\ln (1+x)$  is ${\displaystyle \frac{1}{1+x}}$, which can be represented as the sum of geometric series:

Integrating both from 0 to $x$ gives

(1)

which is valid on the whole open interval of convergence    of this power series and in for  $x=1$, as one may prove.

Replacing $x$ with $-x$ in (1) yields the series

(2)

Subtracting (2) from (1) gives

$\ln {\displaystyle \frac{1+x}{1-x}}=\mathrm{\hspace{0.33em}2}\left(x+{\displaystyle \frac{x^3}{3}}+{\displaystyle \frac{x^5}{5}}+{\displaystyle \frac{x^7}{7}}+\mathrm{\dots }\right)$(3)

which also is true for  .  Here the inner function of the logarithm attains all positive real values when  (its graph (http://planetmath.org/Graph2) is a hyperbola (http://planetmath.org/Hyperbola2) with asymptotes (http://planetmath.org/AsymptotesOfGraphOfRationalFunction)  $x=1$  and  $y=-1$).  Thus, in principle, the series (3) can be used for calculating any values of natural logarithm (http://planetmath.org/NaturalLogarithm2).  For this purpose, one could denote

$$\frac{1+x}{1-x}:=t,$$

which implies

$$x=\frac{t-1}{t+1},$$

and accordingly

$\ln t=\mathrm{\hspace{0.33em}2}\left[{\displaystyle \frac{t-1}{t+1}}+{\displaystyle \frac{1}{3}}{\left({\displaystyle \frac{t-1}{t+1}}\right)}^3+{\displaystyle \frac{1}{5}}{\left({\displaystyle \frac{t-1}{t+1}}\right)}^5+\mathrm{\dots }\right].$(4)

For example,

$$\ln 3=\mathrm{\hspace{0.33em}2}\left(\frac{1}{2}+\frac{1}{3\cdot 2^3}+\frac{1}{5\cdot 2^5}+\mathrm{\dots }\right).$$

The convergence of (4) is the slower the greater is $t$.