HMMT February

Algebra Difficulty 5.1 AIME, harder Find the answer

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 $\frac{1}{A}$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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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