19. [11] Let
$$
A=\lim _{n \rightarrow \infty} \sum_{i=0}^{2016}(-1)^{i} \cdot \frac{\binom{n}{i}\binom{n}{i+2}}{\binom{n}{i+1}^{2}}
$$

Find the largest integer less than or equal to $\frac{1}{A}$.
The following decimal approximation might be useful: $0.6931<\ln (2)<0.6932$, where $\ln$ denotes the natural logarithm function.
Answer:
1

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 .