Hi all Suppose that [itex]a_1=\sqrt5[/itex], [itex]a_{n+1}=a_n^2-2[/itex] and [itex]g_n=\frac{a_1a_2...a_n}{a_{n+1}}[/itex]. Evaluate [itex]\lim_{n\rightarrow \infty } g_n[/itex]. I have seen some information in this link. Besides, the sequence g_{n} seems as a good rational approximation for [itex]\sqrt5[/itex]. I know that the answer is 1, But I can't find any nice solution. Any hint is strongly appreciated.
I have yet to study recurrence relations for myself, but I played around with it for a bit and it seems like it was headed in the right direction. [tex]g_n=\frac{a_1a_2...a_n}{a_n^2-2}[/tex] [tex]=\frac{a_n^2\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)-2\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)+2\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)}{a_n^2-2}[/tex] [tex]=\frac{\left(a_n^2-2\right)\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)+2\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)}{a_n^2-2}[/tex] [tex]=\frac{a_1a_2...a_{n-1}}{a_n}+2\frac{a_1a_2...a_{n-1}}{a_na_{n+1}}[/tex] [tex]=g_{n-1}+\frac{2}{a_n^2}g_n[/tex] Hence we can express g_{n} in terms of g_{n-1}, and the factor multiplying g_{n-1} seems to say a lot about what g_{n} is as n gets very large.
Thanks. I can't see what does it say about g_{n}. Can you help me on this? I found out how to show that: [tex]a_n=\frac{\varphi ^{2^n}-\varphi ^{-2^n}}{\varphi ^{2^{n-1}}-\varphi ^{-2^{n-1}}}[/tex] And then the rest is easy. But I'm still looking for a more elegant solution.
Hmm, after looking at it a bit more I realize it's really not telling us as much as I originally thought. After solving for g_{n} in [tex]g_n=g_{n-1}+\frac{2}{a^2_n-2}g_n[/tex] to obtain [tex]g_n=\frac{a_n^2}{a_n^2-2}g_{n-1}[/tex] I was simply arguing that the factor approaches 1 as n gets large, but this doesn't really tell us about the limit of g_{n}...
hi asmani! reverse-engineering that expression, we get a_{n} = sinh(C2^{n})/sinh(C2^{n-1}) = 2cosh(C2^{n-1}) sooo … try defining a_{n} = 2coshb_{n}
Thanks. I think it's better not to simplify the fraction to get: [tex]g_n=a_1a_2\cdots a_n\frac{1}{a_{n+1}}=\frac{\sinh(C2^1) }{\sinh(C2^0)}\frac{\sinh(C2^2) }{\sinh(C2^1)}\cdots \frac{\sinh(C2^n) }{\sinh(C2^{n-1})}\frac{\sinh(C2^n) }{\sinh(C2^{n+1})}[/tex] [tex]=\frac{\sinh^2(C2^n) }{\sinh(C2^0)\sinh(C2^{n+1})}[/tex] Where C=Log(φ). And now calculating the limit is easy.