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.
 A: $\small \begin{eqnarray} \ln(n) - \ln(n-1) &=&\ln(n)-\left( \ln(n)+\ln({n-1 \over n}) \right) \\ 
 &=& -\ln(1-1/n ) \\ 
&=& 1/n + 1/n^2/2+1/n^3/3+... \end{eqnarray} $     
The latter approximates $\small 1 / n $ when n increases without bounds.
A: \begin{align*}
\lim_{x\to\infty}\frac{\ln x-\ln(x-1)}{1/x} &= \lim_{x\to\infty}x(\ln x-\ln(x-1))\\
&=\lim_{x\to\infty} x\ln\left(\frac{x}{x-1}\right)\\
&=\lim_{x\to\infty}\ln\left(\left(\frac x{x-1}\right)^x\right)\\
&=\ln e\\
&= 1.
\end{align*}
Where $\lim_{x\to\infty}\left(\frac x{x-1}\right)^x=\lim_{x\to\infty}\left(1+\frac1{x-1}\right)^{x-1}\left(1+\frac 1{x-1}\right)=e\cdot 1=e$.
A: You can get quite far with just algebra:
$$\begin{align}
\ln N - \ln (N-1) & = \ln N - ( \ln N + \ln (1-1/N)) \\
& = -\ln (1-1/N)
\end{align}$$
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 $O(1/N^2)$.
A: The two sides of your equation are never exactly equal. But their ratio tends to 1 as $N$ tends to infinity. This is because the derivative of the $\ln$ function at $N$ is $1/N$, so that is approximately the amount by which the function changes between $N-1$ and $N$.
A: Just for completeness, the exact value (I'm surprised it hasn't been mentioned so far, though it's implicit in the "it's the derivative" answer):
$$\begin{align}
\ln N &= \int_{1}^{N} \frac1x \, dx \quad \text{ and }\\
\\
\ln (N+1) &= \int_{1}^{N+1} \frac1x \, dx, \quad \text{ so }\\
\\
\ln (N+1) - \ln N &= \int_{N}^{N+1} \frac1x \, dx
\end{align}$$
Now, as $\frac1{N+1} \le \frac1x \le \frac1N$ for $N \le x \le N+1$, clearly we have
$$ \frac{1}{N+1} \le \ln(N+1) - \ln N \le \frac1{N}$$
Calculating the integral more precisely would give a more precise estimate of $\ln(N+1) - \ln N$.
A: $\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 .
