I have come across a general BBP type formula to calculate the natural log of any integer greater than one. How do I prove that it is true? $$\ln(n) = \sum_{k=1}^\infty \left( n^{-k} \cdot\frac{(n-1)^k}{k} \right) \text{when } n \gt 0$$
The digit-extraction formula is: $$s(x,n) = \sum_{k=1}^{n} \left( \frac{(x-1)^k \cdot x^{n-k} \mod k}{k} \right) + \sum_{k=n+1}^\infty \left( \frac{(x-1)^k \cdot x^{n-k}}{k} \right)$$ $$ts(x,n) = \left\lfloor (s(x,n) \mod 1 ) \cdot x \right\rfloor$$ when $\ln(x)$, and $n + 1$ is the the digit you are trying to find. Generates digits in base $x$.
For $|x|<1$ we have $\ln (1-x)=-\sum_{k=1}^{\infty}x^k/k.$
Let $x=1-1/n.$