Editing Logarithm (section)

====Inverse hyperbolic tangent====
Another series is based on the [[area hyperbolic tangent#Inverse hyperbolic tangent|inverse hyperbolic tangent]] function:
<math display="block">
\ln (z) = 2\cdot\operatorname{artanh}\,\frac{z-1}{z+1} = 2 \left ( \frac{z-1}{z+1} + \frac{1}{3}{\left(\frac{z-1}{z+1}\right)}^3 + \frac{1}{5}{\left(\frac{z-1}{z+1}\right)}^5 + \cdots \right ),
</math>
for any real number {{math|''z'' > 0}}.{{refn|The same series holds for the principal value of the complex logarithm for complex numbers {{mvar|z}} with positive real part.|group=nb}}<ref name=AbramowitzStegunp.68 /> Using [[sigma notation]], this is also written as
<math display="block">\ln (z) = 2\sum_{k=0}^\infty\frac{1}{2k+1}\left(\frac{z-1}{z+1}\right)^{2k+1}.</math>
This series can be derived from the above Taylor series. It converges quicker than the Taylor series, especially if {{mvar|z}} is close to 1. For example, for {{math|1=''z'' = 1.5}}, the first three terms of the second series approximate {{math|ln(1.5)}} with an error of about {{val|3|e=-6}}. The quick convergence for {{mvar|z}} close to 1 can be taken advantage of in the following way: given a low-accuracy approximation {{math|''y'' ≈ ln(''z'')}} and putting
<math display="block">A = \frac z{\exp(y)},</math>
the logarithm of {{mvar|z}} is:
<math display="block">\ln (z)=y+\ln (A).</math>
The better the initial approximation {{mvar|y}} is, the closer {{mvar|A}} is to 1, so its logarithm can be calculated efficiently. {{mvar|A}} can be calculated using the [[exponential function|exponential series]], which converges quickly provided {{mvar|y}} is not too large. Calculating the logarithm of larger {{mvar|z}} can be reduced to smaller values of {{mvar|z}} by writing {{math|''z'' {{=}} ''a'' · 10<sup>''b''</sup>}}, so that {{math|ln(''z'') {{=}} ln(''a'') + {{mvar|b}} · ln(10)}}.

A closely related method can be used to compute the logarithm of integers. Putting <math>\textstyle z=\frac{n+1}{n}</math> in the above series, it follows that:
<math display="block">\ln (n+1) = \ln(n) + 2\sum_{k=0}^\infty\frac{1}{2k+1}\left(\frac{1}{2 n+1}\right)^{2k+1}.</math>
If the logarithm of a large integer&nbsp;{{mvar|n}} is known, then this series yields a fast converging series for {{math|log(''n''+1)}}, with a [[rate of convergence]] of <math display="inline">\left(\frac{1}{2 n+1}\right)^{2}</math>.