evaluate $\lim\limits_{N\to\infty} \dfrac{\ln^2 N}{N}\sum_{k=2}^{N-2} \dfrac{1}{\ln k \cdot \ln (N-k)}$ 
Evaluate $\lim\limits_{N\to\infty} \dfrac{\ln^2 N}{N}\sum_{k=2}^{N-2} \dfrac{1}{\ln k \cdot \ln (N-k)}$ where $\ln$ denotes the natural logarithm.

Let $A_N = \dfrac{\ln^2 N}{N}\sum_{k=2}^{N-2} \dfrac{1}{\ln k \cdot \ln (N-k)}.$ Clearly $A_N \ge \dfrac{\ln^2 N}N \cdot \dfrac{N-3}{\ln^2 N} = 1-3/N$ for all $N$ so the main question is whether $A_N$ converges to 1 as $N\to\infty.$ Fix $2\leq M < N/2.$ We want to find an upper bound for $A_N$ in terms of $N$ and $M$ that converges to 1 as $N\to\infty$ when $M$ is chosen carefully enough. Note that by differentiating $f(x)=\dfrac{1}{\ln x\cdot \ln (N-x)},$ one can conclude that it is decreasing on $(1,N/2]$ and increasing on $[N/2,N-2]$ (one can equivalently analyze the behaviour of $\dfrac{1}{f(x)}$ by differentiating to simplify the calculations).  Using this observation, we have $\sum_{k=2}^{N-2} \dfrac{1}{\ln k\cdot \ln(N-k)} = (\sum_{k=2}^{M} + \sum_{k=M+1}^{N-M-1} + \sum_{k=N-M}^{N-2} )\dfrac{1}{\ln k\cdot \ln(N-k)}\leq \dfrac{2(M-1)}{\ln 2 \ln (N-2)} +\sum_{k=M+1}^{N-M-1} \dfrac{1}{\ln k\cdot \ln(N-k)}.$
But I'm not sure how to simplify the above sum.

Assume that $N > 4$. Let
$$A_N = \frac{\ln^2 N}{N}\sum_{k=2}^{N-2} \frac{1}{\ln k \cdot \ln(N - k)}.$$
Let $f(x) := \frac{1}{\ln x \ln (N - x)}$. Note that $f(x)$ is strictly decreasing on $[2, N/2]$,
and strictly increasing on $[N/2, N-2]$. We have
\begin{align*}
 \sum_{k=2}^{N-2} \frac{1}{\ln k \cdot \ln(N - k)}
 &\le 
 2f(2) + 2\int_2^{N/2} f(x)\,\mathrm{d} x\\
 &\le 2f(2) + 2\int_2^{N/2} \frac{1}{\ln x \ln (N - N/2)}\,\mathrm{d} x\\
 &= 2f(2) + \frac{2}{\ln(N/2)}\int_2^{N/2} \frac{1}{\ln x}\, \mathrm{d} x.
\end{align*}
Also, we have
$$\sum_{k=2}^{N-2} \frac{1}{\ln k \cdot \ln(N - k)}
\ge \sum_{k=2}^{N-2} \frac{1}{\ln N \cdot \ln N} = \frac{N-3}{\ln^2 N}.$$
Thus, we have
$$1 - \frac3N \le A_N
\le 2f(2)\frac{\ln^2 N}{N} + \frac{2\ln^2 N}{N\ln(N/2)}\int_2^{N/2} \frac{1}{\ln x}\, \mathrm{d} x.$$
We have
$$\lim_{N\to \infty} 2f(2)\frac{\ln^2 N}{N} = 0$$
and (using L'Hopital rule)
\begin{align*}
 \lim_{N\to \infty} \frac{2\ln^2 N}{N\ln(N/2)}\int_2^{N/2} \frac{1}{\ln x}\, \mathrm{d} x
 = \lim_{N \to \infty}
 \frac{2\int_2^{N/2} \frac{1}{\ln x}\, \mathrm{d} x}{\frac{N}{\ln N}} = 1.
\end{align*}
We are done.
