Computing $\lim\limits_{n\to \infty} \frac{\ln^2n}{n}\sum\limits_{k=2}^{n-2}\frac{1}{\ln k\ln(n-k)}$

Question

Computing the following limit $$\lim_{n\to \infty} \frac{\ln^2n}{n}\sum_{k=2}^{n-2}\frac{1}{\ln k\ln(n-k)}$$

I first tried to write it as a Riemann sum as follows

$$\frac{\ln^2n}{n}\sum_{k=2}^{n-2}\frac{1}{\ln k\ln(1-\frac{k}{n}) +\ln k\ln n}$$

But this seems not wise. Then tried squeeze the sum as follows

$$ \frac{\ln^2n}{n}\sum_{k=2}^{n-2}\frac{1}{\ln k\ln(n-k)}\ge \frac{\ln^2n}{n}\frac{n-3}{\ln^2n} =1-\frac{3}{n}\to1$$

since $$\ln k\ln(n-k)\le \ln n\ln n=\ln^2 n$$

Therefore I have a feeling that the above searched limit is $1$.

Can any one help to squeeze the sum from above ?

Answer 1

Here is a one way to see this exercise and provide the upper estimate>

I have already proved that $$\frac{\ln^2n}{n}\sum_{k=2}^{n-2}\frac{1}{\ln k\ln(n-k)}\ge1-\frac{3}{n}\to1$$

Now let $2\le m\le n/2$ then since $k\mapsto \frac{1}{\ln k\ln(n-k)}$ is decreasing in $[2,n/2]$ and symmetric with respect to $n/2$ we get

$$\begin{align} A_n &=\frac{\ln^2n}{n}\left\lbrace \sum_{k=2}^{m}+\sum_{k=m+1}^{n-m-1}+\sum_{k=n-m}^{n-2}\right\rbrace \frac{1}{\ln k\ln(n-k)}\\&\le \frac{\ln^2n}{n}\left\lbrace \frac{2(m-1)}{\ln 2\ln(n-2)}+\frac{n-2m-1}{\ln n\ln(n-m)}\right\rbrace \\&\le \frac2{\ln2} \frac{m\ln n}{n}+\left(1-\frac{2m}{n}\right)\frac{\ln n}{\ln m}+O(\frac{1}{\ln n})\end{align}$$

Choose $m= \left\lfloor\frac{n}{\ln^2 n}\right\rfloor+1$ to get,

$$A_n\le \left(1-\frac{2}{n\ln n}\right)\frac{\ln n}{\ln n-2\ln\ln n} ++O(\frac{1}{\ln n}) \to 1$$ at the end of the day we have

$$\lim_{n\to\infty} = \frac{\ln^2n}{n}\sum_{k=2}^{n-2}\frac{1}{\ln k\ln(n-k)}=1$$

Answer 2

$$\frac{\log^2 n}{n}\sum_{k=2}^{n-2}\frac{1}{\log(k)\log(n-k)}=\frac{1}{n}\sum_{k=2}^{n-2}\frac{1}{\left(1+\frac{1}{\log n}\log\frac{k}{n}\right)\left(1+\frac{1}{\log n}\log(1-\frac{k}{n})\right)}$$ where the RHS is a Riemann sum for $$\int_{2/n}^{1-2/n}\frac{dx}{\left(1+\frac{\log x}{\log n}\right)\left(1+\frac{\log(1-x)}{\log n}\right)}.$$ In order to prove that the limit is one we just need to invoke the dominated convergence theorem applied to the last integral or directly to the original series. A squeezing approach can be performed by proving that for any $x\in(0,1)$ we have $$ 4\log^2(2)\,x(1-x)\leq\log(x)\log(1-x) \leq \sqrt{x(1-x)}. $$