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?
 A: This is rather similar to previous answers, but I think it's still worth pointing out. You're asking about the slope of a chord of the graph of $\log x$, the chord joining $(n,\log n)$ to $(n+1,\log(n+1))$. By the mean value theorem, this equals the slope of the tangent line, $1/x$, at some $x$ between $n$ and $n+1$.  
A: 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 $O\left(\frac{1}{n^9}\right)$.
A: Here's a geometric way to get a good approximation for $\ln(n+1)-\ln(n)$:
Use the fact that $\ln(x)=\int_1^x \frac1t dt$.  (For $x>0$.)
Then $\ln(n+1)-\ln(n)$ is the area under the curve $f(x)=\frac1x$ from $n$ to $n+1$.
We are looking for the area over an interval of length $1$.  So numerically the area should be equal to the 'average' height.  Because the $\frac1x$ function is strictly decreasing, we get a good approximation to that 'average' height by evaluating the $\frac1x$ function a the midpoint of the interval (namely at $n+\frac12$).
So $\ln(n+1)-\ln(n)\approx \frac{1}{n+\frac12}$, with the approximation getting better and better as $n$ gets large.
A: $\log(n+1)-\log n=\log(1+\frac1n)$. Using the Taylor series for $\log(1+x)$, this is $$\frac1n-\frac1{2n^2}+\frac1{3n^3}-\cdots\approx\frac1n.$$
A: Especially Lime's answer is absolutely correct and a very good approach. Let me show a conceptually somewhat simpler one, that only uses the derivative, namely $(\log x)'=1/x$. This is a strictly decreasing function, so $\log x$ is concave. In particular, its graph is below the tangent line  at any point of the graph. 
We obtain $\log (n+1)< \log n + 1/n$ and $\log n< \log (n+1) - 1/(n+1)$. To sum up: 
$$\frac{1}{n+1} < \log (n+1)-\log n < \frac{1}{n}$$
