As it is well known, if $|x|<1$ then we can compute $\log(1+x)$ by the Taylor series $$\log(1+x)=x-\frac{x^2}2+\frac{x^3}3-\cdots.$$ Thus, to compute $\log n$ with $n>1$, we may employ the series $$\log n=-\log\left(1-\frac{n-1}n\right)=\sum_{k=1}^\infty\frac{1}k\left(\frac{n-1}n\right)^k,$$ which converges at geometric rate with ratio $(n-1)/n$. Wikipedia provides a more efficient series for computing $\log n$: $$\log n=2\sum_{k=0}^\infty\frac1{2k+1}\left(\frac{n-1}{n+1}\right)^{2k+1}$$ which converges at geometric rate with ratio $(n-1)^2/(n+1)^2$.

For $1<n\le 85/4$, I have found series for $\log n$ which converges at geometric rate with ratio $$-\frac{(n-1)^4}{16n(n+1)^2}.\tag{1}$$ If $1<n<(2+\sqrt5)^2\approx 17.944$, then $$\frac{(n-1)^4}{16n(n+1)^2}<\frac{(n-1)^2}{(n+1)^2}$$ and so my series for computing $\log n$ is more efficient.

**Question.** What's the fastest way to compute $\log n$ for $n>1$? Is there a series for $\log n$ which converges at geometric rate with ratio better than $(1)$?