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 $(1)$?



What's the fastest way to compute $\log n$ for $n>1$?
Theorem 9.1 by [Brent](https://arxiv.org/abs/1004.3412) states the following:



> 
> If $x>0$ is a precision $n$ number, then $\log(x)$ may be evaluated to precision $n$ in time
> $\sim13M(n) \log\_2 n$ as $n\to\infty$ [assuming $\pi$ and $\log(2)$ precomputed to precision $n+O(n/ \log(n))$].
> 
> 
> 


Here
$$M(n)=O(n\log(n)\log\log(n))$$
is the time required to perform a precision $n$ multiplication.


The corresponding method of the evaluation of $\log(x)$ involves A–G mean iterations. Brent also says "There are several algorithms for evaluating $\log(x)$ to precision $n$ in time $O(M(n) \log(n))$."


So, the time to compute $\log(x)$ to precision $n$ is greater than the time to do a precision $n$ multiplication only by a logarithmic factor.


