IMC 1994 · Day 2 · P12

Obviously $$\tag{1} A_N = \frac{\ln^2 N}{N} \sum_{k=2}^{N-2} \frac{1}{\ln k \cdot \ln (N-k)} \ge \frac{\ln^2 N}{N} \cdot \frac{N-3}{\ln^2 N} = 1 - \frac{3}{N}.$$ Take $M$, $2 \le M < N/2$. Then using that $\dfrac{1}{\ln k \cdot \ln (N-k)}$ is decreasing in $[2, N/2]$ and the symmetry with respect to $N/2$ one get $$\begin{align*} A_N &= \frac{\ln^2 N}{N} \left\{ \sum_{k=2}^{M} + \sum_{k=M+1}^{N-M-1} + \sum_{k=N-M}^{N-2} \right\} \frac{1}{\ln k \cdot \ln (N-k)} \le \\ &\le \frac{\ln^2 N}{N} \left\{ 2 \frac{M-1}{\ln 2 \cdot \ln (N-2)} + \frac{N - 2M - 1}{\ln M \cdot \ln (N-M)} \right\} \le \\ &\le \frac{2}{\ln 2} \cdot \frac{M \ln N}{N} + \left( 1 - \frac{2M}{N} \right) \frac{\ln N}{\ln M} + O \left( \frac{1}{\ln N} \right). \end{align*}$$ Choose $M = \left[ \dfrac{N}{\ln^2 N} \right] + 1$ to get $$\tag{2} A_N \le \left( 1 - \frac{2}{N \ln^2 N} \right) \frac{\ln N}{\ln N - 2 \ln\ln N} + O \left( \frac{1}{\ln N} \right) \le 1 + O \left( \frac{\ln\ln N}{\ln N} \right).$$ Estimates (1) and (2) give $$\lim_{N \to \infty} \frac{\ln^2 N}{N} \sum_{k=2}^{N-2} \frac{1}{\ln k \cdot \ln (N-k)} = 1.$$