Solution:
Note
$$\sum_{i=0}^{2016} (-1)^i \cdot \frac{\binom{n}{i} \binom{n}{i+2}}{\binom{n}{i+1}^2} = \sum_{i=0}^{2016} (-1)^i \cdot \frac{(i+1)(n-i-1)}{(i+2)(n-i)}$$
So
$$\lim_{n \rightarrow \infty} \sum_{i=0}^{2016} (-1)^i \cdot \frac{\binom{n}{i} \binom{n}{i+2}}{\binom{n}{i+1}^2} = \sum_{i=0}^{2016} (-1)^i \cdot \frac{(i+1)}{(i+2)} = 1 - \sum_{i=2}^{2016} \frac{(-1)^i}{i} \approx \ln(2)$$
Then $\frac{1}{A} \approx \frac{1}{\ln(2)} \approx 1.44$, so the answer is $1$ .