# Series expansion for $\ln n$ in terms of logs?

Reading Gamma by Julian Havil (really good!) on p. 111 there is a derivation which contains an identity that has me confounded. Where $$n \in \mathbb{N^+}$$, he uses this as one step in the development:

$$\ln n = \sum_{r=2}^n \ln \left( \frac{r}{r-1} \right)$$

I know the basic identities:

$$\ln (1+x) = x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+\cdots =\sum_{r=1}^\infty (-1)^{r-1} \frac{x^r}{r} \qquad -1

and

$$\ln x = \sum_{r=1}^\infty \frac{ (−1)^{r-1}(x-1)^r}{r} \qquad 0

but I can't seem to understand how the author takes that identity above as obvious? What am I missing please?

• Heh, that's actually pretty amusing. Remember that the sum of the logs is the log of the product. So taking the antilog of both sides, you get $n=(2/1)(3/2)(4/3)\cdots(n/(n-1))$. Commented Sep 18, 2022 at 9:57
• Not sure there's ASCII for embarrassing, but thank you @EricSnyder! Commented Sep 18, 2022 at 13:11
• No worries, we all do it sometimes. I guess I should have made it an answer instead for that sweet sweet karma though! /sarcasm Commented Sep 19, 2022 at 9:01

This is because $$n$$ in the formulae is finite, therefore $$\sum_{r = 2}^n\log\left(\frac{r}{r-1}\right) = \log\left(\frac{2}{2 - 1}\right) + \log\left( \frac{3}{3 - 1}\right) + \ldots + \log\left(\frac{n}{n -1}\right) = \log\left(\frac{2}{1}\cdot \frac{3}{2}\cdot\ldots\cdot \frac{n}{n-1}\right) = \log n.$$
Series expansion is not required here. The identity follows from the fact that $$\ln\left(r\over r-1\right)=\ln r-\ln(r-1).$$
• Sure, but that's the substitution he makes, and for a good reason as it turns out later in the derivation as he links the zeta function to $\gamma$. Commented Sep 18, 2022 at 13:07
• I suppose the author's motivation to rewrite $\ln n$ like this is only to prove the existence of that limit for $\gamma$. Commented Sep 18, 2022 at 20:51
The key identity used is the telescopic series: $$\log n = \sum_{r=2}^n (\log r - \log(r-1))$$ (using the fact that $$\log 1 = 0$$ to dismiss the $$\log 1$$). Combine this with $$\log a - \log b = \log\frac{a}{b}$$ to conclude.