# 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.

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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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+1 for nice approach –  Mathlover Apr 16 '12 at 11:17
@Mathlover This was used by Euler when studying $\gamma$ and Harmonic numbers. He saw that $\log \left(1+\dfrac 1 n \right)\sim H_n$ –  Pedro Tamaroff Apr 16 '12 at 18:29

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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And the increment 1 is small when $N$ is large. –  lhf Apr 16 '12 at 11:37
@TonyK This is a nice approach. Kudos! –  Pedro Tamaroff Apr 16 '12 at 18:19
Also, think about $\lim_{N\to\infty}\left(ln N - ln(N-1)\right)$. It's 0, which is also $\lim_{N\to\infty}\left(\frac{1}{N}\right)$. –  chharvey Apr 25 '12 at 4:24

$\small \begin{eqnarray} \ln(n) - \ln(n-1) &=&\ln(n)-\left( \ln(n)+\ln({n-1 \over n}) \right) \\ &=& -\ln(1-1/n ) \\ &=& 1/n + 1/n^2/2+1/n^3/3+... \end{eqnarray}$

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

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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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For example, 1/x^2 does not look like (tend to) 1/x as x->infinity, even though they both tend to 0. However 1/x - 1/x^2 -> 1/x as x-> infinity. So there are two things going on here. Indeed lnN - ln(N-1) -> 1/N as N -> infinity (which is (different, and) stronger than just saying both sides tend to 0) –  Adam Rubinson Apr 16 '12 at 10:57
@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, so the site behavior you experienced was normal. –  Zev Chonoles Apr 16 '12 at 15:05