May I ask why is $\ln N - \ln(N-1) = \frac1N$ for large $N$?
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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 $N$. |
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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 $O(1/N^2)$. |
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$\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. |
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\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 $\lim_{x\to\infty}\left(\frac x{x-1}\right)^x=\lim_{x\to\infty}\left(1+\frac1{x-1}\right)^{x-1}\left(1+\frac 1{x-1}\right)=e\cdot 1=e$. |
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$\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 . |
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