<s>[INST] Limit of recursive sequence Given non-zero sequence $r_n$ such $\lim_{n \to \infty}{r_{2n}} = \frac{1}{2008}$ and $\lim_{n \to \infty}{r_{2n + 1}} = \frac{1}{2009}$ and sequence $a_n$ such $a_1 = x$ and $a_{n+1} = a_n \cdot r_n$ where $x \in \mathbb{R}$. I have to find limit of $a_n$. I noted that $a_n = x^n \cdot r_1 \cdot r_2 \dots r_{n-1}$ but I don't know what to do with it.
 [/INST] Pick $N \in \mathbb{N}$ such that for $n \geq N$,
$$ |r_{2n} - \frac{1}{2008}| \leq \frac{1}{4} $$
$$ |r_{2n+1} - \frac{1}{2009}| \leq \frac{1}{4} $$
which implies
$$ |r_{2n}| \leq \frac{1}{4} + \frac{1}{2008} < \frac{1}{2}$$
$$ |r_{2n+1}| \leq \frac{1}{4} + \frac{1}{2009} < \frac{1}{2}$$
Hence for $n \geq 2N$ we have $|r_n| < \frac{1}{2}$ and therefore
$$|a_n| < |x| \cdot |r_1| \cdot |r_2| \cdot \ldots \cdot |r_{2N}| \cdot \frac{1}{2^{n-2N}}$$ 
Letting $n \rightarrow \infty$ the right hand side of the inequality goes to $0$. Therefore by squeeze theorem,
$$\lim_{n \rightarrow \infty} a_n = 0$$
Remark: You can replace $\frac{1}{4}$ by any $\epsilon > 0$ such that $\epsilon + \frac{1}{2008} < 1$.
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