
#1
Aug811, 02:24 AM

P: 84

Hi all
Suppose that [itex]a_1=\sqrt5[/itex], [itex]a_{n+1}=a_n^22[/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. 



#2
Aug811, 03:07 AM

HW Helper
P: 3,436

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^22}[/tex] [tex]=\frac{a_n^2\left(\frac{a_1a_2...a_{n1}}{a_n}\right)2\left(\frac{a_1a_2...a_{n1}}{a_n}\right)+2\left(\frac{a_1a_2...a_{n1}}{a_n}\right)}{a_n^22}[/tex] [tex]=\frac{\left(a_n^22\right)\left(\frac{a_1a_2...a_{n1}}{a_n}\right)+2\left(\frac{a_1a_2...a_{n1}}{a_n}\right)}{a_n^22}[/tex] [tex]=\frac{a_1a_2...a_{n1}}{a_n}+2\frac{a_1a_2...a_{n1}}{a_na_{n+1}}[/tex] [tex]=g_{n1}+\frac{2}{a_n^2}g_n[/tex] Hence we can express g_{n} in terms of g_{n1}, and the factor multiplying g_{n1} seems to say a lot about what g_{n} is as n gets very large. 



#3
Aug811, 04:07 AM

P: 84

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^{n1}}\varphi ^{2^{n1}}}[/tex] And then the rest is easy. But I'm still looking for a more elegant solution. 



#4
Aug811, 06:05 AM

HW Helper
P: 3,436

A recurrence relation
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_{n1}+\frac{2}{a^2_n2}g_n[/tex] to obtain [tex]g_n=\frac{a_n^2}{a_n^22}g_{n1}[/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}... 



#5
Aug811, 07:03 AM

Sci Advisor
HW Helper
Thanks
P: 26,167

hi asmani!
a_{n} = sinh(C2^{n})/sinh(C2^{n1}) = 2cosh(C2^{n1})sooo … try defining a_{n} = 2coshb_{n} 



#6
Aug811, 08:45 AM

P: 84

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^{n1})}\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. 



#7
Aug811, 04:23 PM

HW Helper
P: 3,436

Sorry about wasting your time asmani, I should've left it to the big guys




#8
Aug911, 03:44 AM

P: 84

No, I appreciate your participation in this thread.



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