Why is $\ln N - \ln(N-1) = \frac1N$ for large $N$? May I ask why is $\ln N - \ln(N-1) = \frac1N$ for large $N$? 

Thank you very much.
You can get quite far with just algebra:

$$\begin{align}
\ln N - \ln (N-1) &amp; = \ln N - ( \ln N + \ln (1-1/N)) \\
&amp; = -\ln (1-1/N)
\end{align}$$

using the laws for addition of logarithms. Now you can use the Taylor expansion of the natural logarithm:

$$-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots$$

to get

$$-\ln(1-1/N) = \frac{1}{N} + \frac{1}{2N^2} + \cdots$$

so that $\ln N - \ln (N-1)$ is, for large $N$, equal to $1/N$ plus a correction term of order $O(1/N^2)$.
$\displaystyle \lim_{n \to \infty} (\ln n-\ln(n-1))=\displaystyle \lim_{n \to \infty}\left(\ln \frac{n}{n-1}\right)=\ln\left(\displaystyle \lim_{n \to \infty} \frac{n}{n-1}\right)=\ln 1=0 $

$\displaystyle \lim_{n \to \infty} \frac{1}{n}=0$

Hence , for large $n$ both $\ln n-\ln(n-1)$ and $\frac{1}{n}$ tends to zero .