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$\{x_n\}$ is a sequence of real numbers such that $\lim_{ n \to \infty} (x_{2n-1}+x_{2n})=2$ and $\lim_{ n \to \infty} (x_{2n}+x_{2n+1})=3$ then what will be $\lim_{n \to \infty} \frac{x_{2n+1}}{x_{2n}}$?

My attempt: $\{x_n\}$ can't be convergent otherwise the two limits should be equal. I was thinking to do $\lim_{ n \to \infty} (x_{2n-1}+x_{2n}+x_{2n+1})$ and compute it two different ways but I can't as the sequence is not convergent. I need some hint where to start form.

lulu
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    Formatting note: using \lim looks better. And \displaystyle \lim_{x\to 0} gets you the arrow underneath the limit, as in $\displaystyle \lim_{x\to 0} f(x)$, if you prefer that. – lulu Jul 28 '23 at 15:27
  • I use \lim\limits_{x\to 0} to format, $\lim\limits_{x\to 0} f(x)$ – Tim Jul 28 '23 at 15:33
  • Thank you @lulu –  Jul 28 '23 at 15:39
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    Subtract the limits to get $\lim_n(x_{2n+1}-x_{2n-1})=1$. You can also change variable $n\to n+1$ in the first, before subtracting to get $x_{2n+2}-x_{2n}\to-1$. So, $x_{2n+1}\to +\infty$ and $x_{2n}\to-\infty$. Use Stolz-Cesaro to finish. – NDB Jul 28 '23 at 15:57
  • Got it, thanks@NDB –  Jul 28 '23 at 16:24

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