HMMT February

Algebra Difficulty 5.8 AIME, harder Prove it United States

Problem:
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.

Solution

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 $1$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.