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Algebra Difficulty 5.1 AIME, harder Prove it United States

Problem:

The sequence of real numbers x1,x2,x3,…x_{1}, x_{2}, x_{3}, \ldots satisfies lim⁡n→∞(x2n+x2n+1)=315\lim_{n \rightarrow \infty}\left(x_{2 n}+x_{2 n+1}\right)=315 and lim⁡n→∞(x2n+x2n−1)=2003\lim_{n \rightarrow \infty}\left(x_{2 n}+x_{2 n-1}\right)=2003. Evaluate lim⁡n→∞(x2n/x2n+1)\lim_{n \rightarrow \infty}\left(x_{2 n} / x_{2 n+1}\right).

Solution

Solution:

We have lim⁡n→∞(x2n+1−x2n−1)=lim⁡n→∞[(x2n+x2n+1)−(x2n+x2n−1)]=315−2003=−1688\lim_{n \rightarrow \infty}\left(x_{2 n+1}-x_{2 n-1}\right)=\lim_{n \rightarrow \infty}\left[\left(x_{2 n}+x_{2 n+1}\right)-\left(x_{2 n}+x_{2 n-1}\right)\right]=315-2003=-1688; it follows that x2n+1→−∞x_{2 n+1} \rightarrow -\infty as n→∞n \rightarrow \infty. Then

lim⁡n→∞x2nx2n+1=lim⁡n→∞x2n+x2n+1x2n+1−1=−1 \lim_{n \rightarrow \infty} \frac{x_{2 n}}{x_{2 n+1}}=\lim_{n \rightarrow \infty} \frac{x_{2 n}+x_{2 n+1}}{x_{2 n+1}}-1=-1

since x2n+x2n+1→315x_{2 n}+x_{2 n+1} \rightarrow 315 while x2n+1→−∞x_{2 n+1} \rightarrow -\infty.

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