# Evaluating $\lim_{n\to\infty}\sum_{1\leq i<k\leq n}\ln(\sqrt[n]{\frac{3n-2i}{3n+2i}})\ln(\sqrt[n]{\frac{3n-2k}{3n+2k}})$

$$\lim_{n\to\infty}\sum_{1\leq i Okay so first off, what pops off is the sum notation, I've read tons of posts here about the notation when it is presented this way and came to conclusion that this can be also expressed as: $$\lim_{n\to\infty}\sum_{i=1}^n\sum_{k=1}^n\ln\left(\sqrt[n]{\frac{3n-2i}{3n+2i}}\right)\ln\left(\sqrt[n]{\frac{3n-2k}{3n+2k}}\right)$$ if that's wrong, please do let me know. About the actual problem, I'm quite out of ideas, I thought that we could do this: $$\lim_{n\to\infty}\left(\sum_{i=1}^n \ln\left(\sqrt[n]{\frac{3n-2i}{3n+2i}}\right)\sum_{k=1}^n \ln\left(\sqrt[n]{\frac{3n-2k}{3n+2k}}\right)\right)$$ since one is just considered to be constant with the other? Not sure if that's valid. And then I'm thinking this involves some kind of riemann sums and integrals? (Considering the last step is correct, if it's not then no)

• To add to everything else. No your conclusion about the sum is wrong. In this case the answer is half of that. You need to be really careful about the indices of the summation. Sep 11, 2021 at 8:48

You need to use Riemann Sums in 2 dimensions.

$$\frac{\lim_{n\to n}\sum_{i=1}^{n}\sum_{k=1}^{n}\ln\left(\sqrt[n]{\frac{3n-2i}{3n+2i}}\right)\ln\left(\sqrt[n]{\frac{3n-2k}{3n+2k}}\right)-\lim_{n\to\infty}\sum_{i=k}\ln\left(\sqrt[n]{\frac{3n-2i}{3n+2i}}\right)\ln\left(\sqrt[n]{\frac{3n-2k}{3n+2k}}\right)}{2}$$

$$=\frac{\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{k=1}^{n}\ln\left(\sqrt[n]{\frac{3n-2i}{3n+2i}}\right)\ln\left(\sqrt[n]{\frac{3n-2k}{3n+2k}}\right)-\lim_{n\to\infty}\sum_{i=1}^{n}\left(\ln^{2}\left(\sqrt[n]{\frac{3n-2i}{3n+2i}}\right)\right)}{2}$$ Note here that:- $$\lim_{n\to\infty}\sum_{i=1}^{n}\left(\ln^{2}\left(\sqrt[n]{\frac{3n-2i}{3n+2i}}\right)\right)=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\frac{1}{n}\left(\ln^{2}\left(\frac{3n-2i}{3n+2i}\right)\right)\approx\\ \lim_{n\to\infty}\frac{1}{n}\int_{0}^{1}\ln^{2}(\frac{(1-\frac{2x}{3})}{(1+\frac{2x}{3})})dx$$

This is just by Riemann sum of 1 variable.

$$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}f(\frac{i}{n})=\int_{0}^{1}f(x)dx$$

And $$\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{k=1}^{n}\ln\left(\sqrt[n]{\frac{3n-2i}{3n+2i}}\right)\ln\left(\sqrt[n]{\frac{3n-2k}{3n+2k}}\right)=\lim_{n\to\infty}\frac{1}{n^{2}}\sum_{i=1}^{n}\sum_{k=1}^{n}\ln\left(\frac{1-\frac{2i}{3n}}{1+\frac{2i}{3n}}\right)\ln\left(\frac{1-\frac{2k}{3n}}{1+\frac{2k}{3n}}\right)=\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{(1-\frac{2x}{3})}{(1+\frac{2x}{3})}\right)\ln\left(\frac{(1-\frac{2y}{3})}{(1+\frac{2y}{3})}\right)dydx$$

This happens due to $$\lim_{n\to\infty}\frac{1}{n^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n}f(\frac{i}{n},\frac{j}{n})=\int_{0}^{1}\int_{0}^{1}f(x,y)dxdy$$

So we have:-

$$\displaystyle\lim_{n\to\infty}\frac{\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{(1-\frac{2x}{3})}{(1+\frac{2x}{3})}\right)\ln\left(\frac{(1-\frac{2y}{3})}{(1+\frac{2y}{3})}\right)dydx-\frac{1}{n}\int_{0}^{1}\ln^{2}(\frac{(1-\frac{2x}{3})}{(1+\frac{2x}{3})})dx}{2}$$

Now you would have to evaluate this integral. In the 2nd integral what we have done is taken the Riemann Integral but for one variable. Because you only need $$\frac{1}{n}$$ to get the riemann sum for one index. The other $$\frac{1}{n}$$ is in the denominator.

So since the integral is finite, when $$n\to\infty$$. The limit tends to $$0$$.

So you only have $$\displaystyle\frac{\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{(1-\frac{2x}{3})}{(1+\frac{2x}{3})}\right)\ln\left(\frac{(1-\frac{2y}{3})}{(1+\frac{2y}{3})}\right)dydx}{2}=\frac{\left(\int_{0}^{1}\ln\left(\frac{(1-\frac{2x}{3})}{(1+\frac{2x}{3})}\right)dx\right)^{2}}{2}=\frac{(3\ln(3)-\frac{5}{2}\ln(5))^{2}}{2}$$

The iterated integral is just the square of the integral wrt $$x$$ as they are both equal. If you don't prefer the square , then you can just compute wrt y first and you will get value :- $$(3\ln(3)-\frac{5}{2}\ln(5))$$ . And after this you compute it wrt x and you will again get $$(3\ln(3)-\frac{5}{2}\ln(5))$$ . So the value of the integral is $$\left(3\ln(3)-\frac{5}{2}\ln(5)\right)^{2}$$

The integral of $$\int_{0}^{1}\ln(\frac{(1-\frac{2x}{3})}{(1+\frac{2x}{3})})dx$$ is simple enough to compute. Just write the integrand as $$\ln((1-\frac{2x}{3})-\ln(1+\frac{2x}{3}))$$ and then use substitution $$\frac{2x}{3}=t$$. And then use antiderivative of $$\ln(x)$$ as $$x\ln(x)-x$$

More explanation for @acyex . You are familiar with Riemann sums of 1 variable right?. The idea in 2D is the same . Here you can say that $$\lim_{n\to\infty}\frac{1}{n^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n}f(\frac{i}{n},\frac{j}{n})=\int_{0}^{1}\int_{0}^{1}f(x,y)dxdy$$ This is precisely what I have done here.

Now you see that the in the summation that you are given , the indices do not vary independently. For example if I say to you to compute $$\sum_{i=1}^{2}\sum_{j=1}^{2} a_{i}b{j}$$ what would you do?.

You would simply write it as $$a_{1}b_{1}+a_{1}b_{2}+a_{2}b_{1}+a_{2}b_{2}$$ , right? . That is you sum with the j index first keeping i as a constant and then sum over the index i.

But here you cannot do that . $$\displaystyle\sum_{1\leq i as only in this case the index j is greater than that of i. That is precisely what I have done there. As $$\displaystyle\sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{j} = \sum_{1\leq i=j\leq n}a_{i}b_{j} + \sum_{1\leq i. But what the question wants is $$\displaystyle\sum_{1\leq i .

But notice that in this case $$\displaystyle\sum_{1\leq i. This happens due to the fact that $$a_{i}$$ and $$b_{j}$$ are the same $$\ln\left(\sqrt[n]{\frac{3n-2k}{3n+2k}}\right)$$ and $$\ln\left(\sqrt[n]{\frac{3n-2i}{3n+2i}}\right)$$ . So by symmetry $$a_{1}b_{2}=a_{2}b_{1}$$. This is not true in a general case. But here it works out perfectly fine due to both the functions in the summand being symmetric wrt to i and j (or i and k as in the question).

So what we have is $$\displaystyle\sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{j} = \sum_{1\leq i=j\leq n}a_{i}b_{j} + 2\sum_{1\leq i. So $$\displaystyle\sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{j} -\sum_{1\leq i=j\leq n}a_{i}b_{j}=2\sum_{1\leq i

Or , $$\displaystyle \sum_{1\leq i

Now here see that $$\displaystyle\sum_{1\leq i=j\leq n}a_{i}b{j}=\sum_{i=1}^{n}a_{i}b_{i}$$.

Hence we have $$\displaystyle \sum_{1\leq i

I hope it is clear now. I guarantee you no one will explain as much as this.

And now I would like to ask a personal question to you. @Acyex. What grade are you in?. Are you in High school and familiar with integrals and simple riemann sums which lets you view sums as integrals? .Or are you in college where you actually know Riemann Integration and know why this is justified and you know Multivariable calculus?. Or are you completely unfamiliar with Riemann sums?. If yes, then you should skip and avoid this question totally as there is no other way to evaluate this summation without Riemann sums. You can revisit it once you have learnt these things. This is not a question that you find it in "limits" section of a calculus book. This involves integrals. And double integrals at that.

• in the integral the typo is unclear (logarithm) can you improve ? Sep 7, 2021 at 14:08
• I don't understand which line you want me to edit. Can you please specify? Sep 7, 2021 at 14:14
• No need to specify now looks good ! Sep 7, 2021 at 14:23
• @Acyex that notation is a consequence of two common conventions. It is common to (specifically) write $\sum_{i=1}^{n} = sum$, omitting the i=1->n to be assumed by the reader. The remaining part of the notation $i=k, i<k, i>k$ is placing constraints on each part of the sum, and in particular the double sum to reduce it down to a single sum. So the first summand you are confused about $\sum_{i=k}$ is simply saying, sum this equation, over i (and k) =1 to n, with the constraint that i=k. Sep 10, 2021 at 9:45
• Similarly the $i<k$ is a different constraint, that can be written $\sum_{i<k} = \sum_{i=1}^{n} \sum_{k=i+1}^{n}$. @Mr.GandalfSauron is splitting out the more general double sum in this way so he can work on separate parts of it, to arrive at what you actually want to evaluate $\sum_{i<k}$ Sep 10, 2021 at 9:51