Limit of the sequence $\frac{1}{n}\left[\log\left(\frac{n+1}{n}\right)+\log\left(\frac{n+2}{n}\right)+\dots+\log\left(\frac{n+n}{n}\right)\right]$

Question

How can we evaluate the following limit $$ \lim_{n\to\infty}\frac{1}{n}\left[\log\left(\frac{n+1}{n}\right)+\log\left(\frac{n+2}{n}\right)+\dots+\log\left(\frac{n+n}{n}\right)\right] $$

Answer 1

Using the first principle, $$\begin{align} \\ & \lim_{n\to\infty}\frac{1}{n}\left[\log\left(\frac{n+1}{n}\right)+\log\left(\frac{n+2}{n}\right)+\dots+\log\left(\frac{n+n}{n}\right)\right] \\ & =\lim_{n\to\infty}\frac{1}{n}\left[\log\left(1+\frac{1}{n}\right)+\log\left(1+\frac{2}{n}\right)+\dots+\log\left(1+\frac{n}{n}\right)\right] \\ & =\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n}\log\left(1+\frac{r}{n}\right) \\ & =\lim_{h\to 0} h \cdot \sum_{r=1}^{n}\log\left(1+rh\right) \\ & =\int_0^1 \log\left(1+x\right) dx \\ & =\frac{1}{\ln 10}\int_0^1 \ln\left(1+x\right) dx \\ & =\frac{1}{\ln 10}\left[x\ln\left(1+x\right)-x+\ln\left(1+x\right)\right]_0^1 \\ & =\frac{1}{\ln 10}\left[(x+1)\ln\left(1+x\right)-x\right]_0^1 \\ & =\frac{2\ln\left(2\right)-1}{\ln 10} \end{align}$$

Answer 2

The idea is to remark that it's a Riemann sum :

$$\frac{1}{n} \sum_{k=1}^n \ln( 1 + \frac{k}{n} ) $$

So

$$\lim \frac{1}{n} \sum_{k=1}^n \ln( 1 + \frac{k}{n} ) = \int_0^1 \ln(1+x) dx = \left[ (x+1)\ln(1+x) -x \right]_0^1 = 2\ln(2) - 1$$