Limit using Riemann integral Okay so I have to find the following limit of the function given as :
$$\lim_{n\to\infty} \left(\frac{1}{n}\cdot\frac{2}{n}\cdot\frac{3}{n}\cdots\cdots\frac{n}{n}\right)^\frac{1}{n} $$
Now , on taking $log$ both sides and rearranging , I get something like this 
$$ln A =\frac{1}{n} \left[ ln \left(\frac{1}{n}\right)+ln \left(\frac{2}{n}\right)+ln \left(\frac{3}{n}\right)+ \cdot \cdot \cdot \cdot \cdot \cdot \cdot ln \left(\frac{n}{n}\right)\right]$$
OR 
$$ln A = \sum_{r=1}^n \frac{1}{n} \cdot ln \left(\frac{r}{n}\right)$$
 Now I want to convert it into a Riemann sum , but I have no idea how I can do that , also , what would be the limits of that integral ? Can someone please help me on this ? I'm really new to Riemann sums and using this technique to find limits .
 A: That
$$\tag 1 \sum_{k=1}^{n}\frac{1}{n}\ln \left (\frac{k}{n}\right) \to \int_0^1 \ln x\, dx$$
as $n\to \infty$ needs some justification. First, $\ln x$ is unbounded on $(0,1].$ Thus $\ln x$ is not Riemann integrable on $[0,1].$ However, the improper integral $\int_0^1 \ln x\, dx$ converges. That is good to know, but is it always true that a convergent improper integral is the limit of such Riemann-like sums? The answer is no. There are counterexamples. However, the answer is yes if the integrand is monotone. Since $\ln x$ is strictly increasing on $(0,1]$ we can be sure that $(1)$ holds.
A: $$\lim_{n\to\infty}\left\{\sum_{r=1}^n\frac1n\ln\left(\frac rn\right)\right\}=\lim_{n\to\infty}\left\{\int_1^n\frac1n\ln\left(\frac xn\right)\,dx\right\}=\lim_{n\to\infty}\left\{\int_{\frac1n}^1\ln(y)\,dy\right\}=\int_0^1\ln y\,dy$$
This evaluates to $-1$, so $A=e^{-1}$.
A: $$\lim_{n\to\infty}\sum_{r=1}^n \frac{1}{n} \cdot \ln \left(\frac{r}{n}\right)  =\int_0^1 \ln xdx =[x\ln x-x]_0^1 =-1$$
Hence
$$\lim_{n\to\infty} \left(\frac{1}{n}\cdot\frac{2}{n}\cdot\frac{3}{n}\cdots\cdots\frac{n}{n}\right)^\frac{1}{n}=e^{-1}$$
