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

Question

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:

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

and

$$ \ln x = \sum_{r=1}^\infty \frac{ (−1)^{r-1}(x-1)^r}{r} \qquad 0<x\le 2 $$

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

Answer 1

This is because $n$ in the formulae is finite, therefore $$ \sum_{r = 2}^n\log\left(\frac{r}{r-1}\right) = \log\left(\frac{2}{2 - 1}\right) + \log\left( \frac{3}{3 - 1}\right) + \ldots + \log\left(\frac{n}{n -1}\right) = \log\left(\frac{2}{1}\cdot \frac{3}{2}\cdot\ldots\cdot \frac{n}{n-1}\right) = \log n. $$

Answer 2

Series expansion is not required here. The identity follows from the fact that $$ \ln\left(r\over r-1\right)=\ln r-\ln(r-1). $$

Answer 3

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.