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

Thank you very much.

6 Answers 6

11

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 $O(1/N^2)$.

  • 1
    @Mathlover This was used by Euler when studying $\gamma$ and Harmonic numbers. He saw that $\log \left(1+\dfrac 1 n \right)\sim H_n $2012-04-16
10

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 $N$.

  • 0
    Also, think about $\lim_{N\to\infty}\left(ln N - ln(N-1)\right)$. It's 0, which is also $\lim_{N\to\infty}\left(\frac{1}{N}\right)$.2012-04-25
6

$\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.

2

\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 $\lim_{x\to\infty}\left(\frac x{x-1}\right)^x=\lim_{x\to\infty}\left(1+\frac1{x-1}\right)^{x-1}\left(1+\frac 1{x-1}\right)=e\cdot 1=e$.

1

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 $\ln(N+1) - \ln N$.

-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 .

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    @Adam: I just wanted to mention that because you do not have 50 reputation points yet, [you can only comment on your own questions and answers](http://meta.stackexchange.com/questions/19756/how-do-comments-work/19757#19757), so the site behavior you experienced was normal.2012-04-16