What is the limit of the recurrence relation for g_n?

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The limit of the recurrence relation defined by \( g_n = \frac{a_1 a_2 \ldots a_n}{a_{n+1}} \) varies based on the initial value \( a_1 \). If \( a_1 = 0 \), then \( g_n \) equals zero for all \( n \). For \( a_1 = 1 \), the sequence becomes alternating and does not converge. When \( a_1 = 2 \), \( g_n \) diverges to infinity as it simplifies to \( 2^{n-1} \). Therefore, the limit exists only for specific values of \( a_1 \).

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Hi all

Suppose that , a_{n+1}=a_n^2-2 and g_n=\frac{a_1a_2...a_n}{a_{n+1}}.
Evaluate \lim_{n\rightarrow \infty } g_n.

I have seen some information in this link. Besides, the sequence gn seems as a good rational approximation for \sqrt5. I know that the answer is 1, But I can't find any nice solution. Any hint is strongly appreciated.
 
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I'm not sure the limit exists in general ...

If [tex]a_1=0[/tex] then [tex]g_n=0[/tex] for all n, so the limit is zero. If [tex]a_1=1[/tex], then g is an alternating sequence and the limit does not exist. If [tex]a_1=2[/tex], then [tex]g_n=2^{n-1}[/tex] which diverges. In fact, after playing for a few minutes, I can only get g to either converge to zero or not converge at all.

jason
 

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