Recognitions:

## Re: Probability for the first digit of a natural number being equal to 1

 Quote by D H To answer CR's question: Yes, it does, thanks for that! The logarithmic density is ln 2/ln 10 = log10 2 -- and that is exactly what Benford's law says.
How do you show that? It explicitly contradicts what we have discussed.

 Mentor The definition of logarithmic density of some subset $A\in\mathbb N$ is $$\delta(A) \equiv \lim_{n\to\infty} \frac{\,\,\displaystyle {\sum_{\substack{r<=n,\\r\in A}}\frac 1 r}\,\,} {\displaystyle \sum_{r=1}^n\frac 1 r}$$ ... if that limit exists. Alternatively, using lower and upper limits, the logarithmic density is the lower or upper limit if both of those limits exist and are equal. Those two limits are equal in this case. Let f be some real number in [1, 10) (i.e., the mantissa of a real number in base 10). Denote \begin{align*} C_{f\cdot 10^n} &= \sum_{\substack{r<=\lfloor f\cdot10^n \rfloor,\\r\in A}}\frac 1 r \\[4pt] H_{f\cdot 10^n} &=\sum_{r=1}^{\lfloor f\cdot 10^n \rfloor} \frac 1 r \end{align*} Note that used H here because the denominator $H_{f\cdot 10^n}$ is the $\lfloor f\cdot 10^n \rfloor^{\text{th}}$ harmonic number. For large n, \begin{align*} C_{f10^n} &\to (n+\mathcal O(1))\,\ln 2 \\[4pt] H_{f10^n} &\to (n+\mathcal O(1))\,\ln 10 \\[4pt] \end{align*} The mantissa f is absorbed in that O(1) term. In the limit $\n\to\infty$, the ratio becomes $\ln 2/\ln 10$, or log10 2.

Mentor
 Quote by tom.stoer How do you show that? It explicitly contradicts what we have discussed.
I showed that just above. Just because the natural density of a set does not exist does not mean that the logarithmic density does not exist.

 Recognitions: Science Advisor I see; thanks for the explanation. Last question: how do you "guess" which density is reasonable?