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

$$ln A =\frac{1}{n} \left[ ln \left(\frac{1}{n}\right)+ln \left(\frac{2}{n}\right)+ln \left(\frac{3}{n}\right)+ \cdot \cdot \cdot \cdot \cdot \cdot \cdot ln \left(\frac{n}{n}\right)\right]$$


$$ln A = \sum_{r=1}^n \frac{1}{n} \cdot ln \left(\frac{r}{n}\right)$$ 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 .

  • $\begingroup$ It's $\int_0^1 \ln x \ \mathrm{d}x$ $\endgroup$ – Crostul Dec 7 '17 at 17:29
  • $\begingroup$ Exactly ! But can you explain it to me ? I just am not able to see why it is what it is . $\endgroup$ – Tanuj Dec 7 '17 at 17:30
  • $\begingroup$ The basic method is to take $r/n$ as $x$ and $1/n$ as $dx$. The limits of integral are limiting value of $x$ based on extreme values of $r$. $\endgroup$ – samjoe Dec 7 '17 at 17:31
  • $\begingroup$ @samjoe , do I always have to take r/n as x and 1/n as dx ? And I still did not get how to get the limits of the integral. $\endgroup$ – Tanuj Dec 7 '17 at 17:33
  • $\begingroup$ @Tanuj since you are summing over $r$, this is what you would take as your $x$, since you integrate over $x$. If you choose to take $r/n$ as your $x$, then you get $1/n$ as your $dx$ by consequence. You get the limits by this substitution (see my answer) $\endgroup$ – John Doe Dec 7 '17 at 20:19


$$\tag 1 \sum_{k=1}^{n}\frac{1}{n}\ln \left (\frac{k}{n}\right) \to \int_0^1 \ln x\, dx$$

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.

  • $\begingroup$ Will my job be done always by taking r/n as x and then 1/n as dx ? Is this the standard method in this technique or does this apply to only this question ? $\endgroup$ – Tanuj Dec 8 '17 at 3:04

$$\lim_{n\to\infty}\left\{\sum_{r=1}^n\frac1n\ln\left(\frac rn\right)\right\}=\lim_{n\to\infty}\left\{\int_1^n\frac1n\ln\left(\frac xn\right)\,dx\right\}=\lim_{n\to\infty}\left\{\int_{\frac1n}^1\ln(y)\,dy\right\}=\int_0^1\ln y\,dy$$ This evaluates to $-1$, so $A=e^{-1}$.


$$\lim_{n\to\infty}\sum_{r=1}^n \frac{1}{n} \cdot \ln \left(\frac{r}{n}\right) =\int_0^1 \ln xdx =[x\ln x-x]_0^1 =-1$$


$$\lim_{n\to\infty} \left(\frac{1}{n}\cdot\frac{2}{n}\cdot\frac{3}{n}\cdots\cdots\frac{n}{n}\right)^\frac{1}{n}=e^{-1}$$


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