Determine the limit by the condition given

Question

If a real sequence $(x_n)$ satisfies $\displaystyle \lim_{n \to \infty} x_{2n} + x_{2n+1} = 315$ and $\displaystyle \lim_{n \to \infty} x_{2n} + x_{2n-1} = 2016$ then $\displaystyle \lim_{n \to \infty} \dfrac{x_{2n}}{x_{2n+1}} = \cdots$

I really don't know how to approach this question, which come from real analysis olympiad in my country. My friend tried to input $x_{2n}$ as $300$ as its limit, and $x_{2n+1}$ as $15$ so it satisfies the first one, and hence the limit of what the question ask is $20$. But I really don't accept this idea. Could you help me?

Answer 1

$\displaystyle \lim_{n \to \infty} \dfrac{x_{2n}}{x_{2n+1}} \stackrel{by Stolz-Cesaro}= \lim_{n \to \infty} \dfrac{x_{2n}-x_{2n-2}}{x_{2n+1}-x_{2n-1}}=\lim_{n \to \infty} \dfrac{x_{2n}+x_{2n-1}-(x_{2n-2}+x_{2n-1})}{x_{2n+1}+x_{2n}-(x_{2n}+x_{2n-1})}=\frac{2016-315}{315-2016}=-1$.

Answer 2

Hint: Apply Stolz-Cesaro theorem to sequences $a_n=x_{2n}$ (as numerator)and $b_n=x_{2n+1}$ (as denominator). Make sure you first check that $b_n$ is strictly monotonic and unbounded (for large enough $n$).