A fast series for ln 2

1. $\ln(1+t)-\ln(1-t)=2\left(t+\frac{t^3}3+\frac{t^5}5+o(t^5)\right)$.

2. For $t=\frac13$: $\frac{1+t}{1-t}=2$, so
$$\ln2\approx2\left(\frac13+\frac1{81}+\frac1{1215}\right)=0.69300\ldots$$
The full series $2\sum_{k\ge0}\frac{t^{2k+1}}{2k+1}$ has positive terms, and the tail after $k=2$ is bounded by a geometric series:
$$0<\text{error}<2\cdot\frac{t^7}7\cdot\frac1{1-t^2}=\frac2{7\cdot2187}\cdot\frac98<1.5\cdot10^{-4}.$$
So $0.6930\le\ln2\le0.6932$ (true value $0.693147\dots$).

3. The series $\ln(1+x)=x-\frac{x^2}2+\cdots$ at $x=1$ gives the alternating harmonic series, whose error after $n$ terms is of order $\frac1n$: reaching $10^{-4}$ would need about $10^4$ terms, against just $3$ terms above.

Remark. Choosing the expansion variable cleverly ($t=\frac13$ instead of $x=1$) is what makes numerical computations of $\ln$ practical.