<s>[INST] 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?
 [/INST] 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 $O\left(\frac{1}{n^9}\right)$.
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