What is $\log(n+1)-\log(n)$? What is gap $\log(n+1)-\log(n)$ between log of consecutive integers? That is what precision of logarithms determines integers correctly?

Just added for your curiosity.
In the same spirit as in other answers, instead of Taylor series, you could consider Padé approximants and get things such as
$$\log(n+1)-\log n=\log\left(1+\frac1n\right) \approx \frac{2}{2 n+1}$$
$$\log(n+1)-\log n=\log\left(1+\frac1n\right) \approx \frac{6 n+3}{6 n^2+6n+1}$$
$$\log(n+1)-\log n=\log\left(1+\frac1n\right) \approx \frac{60 n^2+60 n+11 } {60 n^3+90 n^2+36 n+3 }$$
$$\log(n+1)-\log n=\log\left(1+\frac1n\right) \approx \frac{420 n^3+630 n^2+260 n+25 }{420 n^4+840 n^3+540 n^2+120 n+6 }$$
These are respectively equivalent to Taylor series to $O\left(\frac{1}{n^3}\right)$, $O\left(\frac{1}{n^5}\right)$, $O\left(\frac{1}{n^7}\right)$ and $O\left(\frac{1}{n^9}\right)$.
