When attempting the evaluate the integral $\int_0^1 \{\ln(x)\}$, where $\{ x \}$ is the fractional part function, I came across the following sum: $$\sum_{r = 1}^n \ln\left(\frac{r + 1}{r}\right) $$ While the sum can be rewritten with the quotient property of the logarithm and telescoped to reach $\ln(n + 1)$, I came across, in a Mathematics MI video, an interesting alternative method. His argument (transcribed verbatim) is as follows: $$ \begin{align} \sum_{r = 1}^n \ln\left(\frac{1 + r}{r}\right) &= \ln (\Gamma(n + 2)) - \ln (\Gamma(n+1)) \\ &= \ln \left( \frac{\Gamma (n+2)}{\Gamma(n+1)} \right) \\ &= \ln \left( n+1 \right) \end{align} $$
Is the first line of the equation valid? If so, where does the identity come from?
Thank you very much!
Proof:
$$ \begin{align}\ln(\Gamma(n+2))-\ln(\Gamma(n+1)) &= \ln\left(\frac{\Gamma(n+2)}{\Gamma(n+1)}\right)\\ &= \ln\left(\frac{(n+1)!}{n!}\right) \\ & = \ln(n+1)\\ &= \displaystyle\sum_{r=1}^{n}\ln\left(\frac{1+r}{r}\right)\end{align} $$ $\blacksquare$