section_path: Calculation/Power series/Inverse hyperbolic tangent
section_name: Inverse hyperbolic tangent
paragraph_text: Inverse hyperbolic tangent
  
  Another series is based on the inverse hyperbolic tangent function: ln (z) = 2 ⋅ artanh z − 1 z + 1 = 2 (z − 1 z + 1 + 1 3 (z − 1 z + 1) 3 + 1 5 (z − 1 z + 1) 5 + ⋯), {\displaystyle \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),} for any real number z > 0. Using sigma notation, this is also written as ln (z) = 2 ∑ k = 0 ∞ 1 2 k + 1 (z − 1 z + 1) 2 k + 1. {\displaystyle \ln(z)=2\sum _{k=0}^{\infty }{\frac {1}{2k+1}}\left({\frac {z-1}{z+1}}\right)^{2k+1}.} This series can be derived from the above Taylor series. It converges quicker than the Taylor series, especially if z is close to 1. For example, for z = 1.5, the first three terms of the second series approximate ln(1.5) with an error of about 3 × 10. The quick convergence for z close to 1 can be taken advantage of in the following way: given a low-accuracy approximation y ≈ ln(z) and putting A = z exp (y), {\displaystyle A={\frac {z}{\exp(y)}},} the logarithm of z is: ln (z) = y + ln (A). {\displaystyle \ln(z)=y+\ln(A).} The better the initial approximation y is, the closer A is to 1, so its logarithm can be calculated efficiently. A can be calculated using the exponential series, which converges quickly provided y is not too large. Calculating the logarithm of larger z can be reduced to smaller values of z by writing z = a · 10, so that ln(z) = ln(a) + b · ln(10).
  
  A closely related method can be used to compute the logarithm of integers. Putting z = n + 1 n {\displaystyle \textstyle z={\frac {n+1}{n}}} in the above series, it follows that: ln (n + 1) = ln (n) + 2 ∑ k = 0 ∞ 1 2 k + 1 (1 2 n + 1) 2 k + 1. {\displaystyle \ln(n+1)=\ln(n)+2\sum _{k=0}^{\infty }{\frac {1}{2k+1}}\left({\frac {1}{2n+1}}\right)^{2k+1}.} If the logarithm of a large integer n is known, then this series yields a fast converging series for log(n +1), with a rate of convergence of (1 2 n + 1) 2 {\textstyle \left({\frac {1}{2n+1}}\right)^{2}}.