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Suppose that $a_0 = \sqrt5$, $a_{n+1} = 2a_n^2-1$. Define $b_n :=2^na_0a_1...a_{n-1}$. Show that $\lim_{n\to\infty}\frac{a_n}{b_n} = 2$.

I can show that this limit exists by showing that the sequence is contractive, after which I have no clue how to find its value. Could someone point me in the right direction? The limit sort of makes sense because both $a_n$ and $b_n$ are more or less doubling in value each time, with the former one 'step' ahead of the latter

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  • $\begingroup$ does it converges to $2$ ?? ,i got a different answer $\endgroup$ – Nebo Alex Apr 21 '16 at 18:03
  • $\begingroup$ I believe it does; managed to derive it after obtaining a hint from someone. Care to share how you got your different answer? $\endgroup$ – Zhanfeng Lim Apr 22 '16 at 5:42
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We have with $a_n=\cosh(u_n)$, $u_{n+1}=2u_n$ giving $u_n=2^nargcosh(a_0)$. So $a_n=\cosh(2^nargcosh(a_0))$ yielding

$$b_n=\prod_{k=0}^{n-1}2a_k=\prod_{k=0}^{n-1}2\cosh(2^kargcosh(a_0))=\prod_{k=0}^{n-1}\dfrac{\sinh(2^{k+1}argcosh(a_0))}{\sinh(2^{k}argcosh(a_0))}=\dfrac{\sinh(2^{n}argcosh(a_0))}{\sinh(argcosh(a_0))}.$$

It follows that $$\dfrac{a_n}{b_n}=\dfrac{\sinh(argcosh(a_0))\cosh(2^nargcosh(a_0))}{\sinh(2^nargcosh(a_0))}=\tanh(2^nargcosh(a_0))\sinh(argcosh(a_0))\underset{n\to+\infty}{\longrightarrow}\sinh(argcosh(a_0)).$$

As $$\sinh(argcosh(a_0))=\sinh(ln(2+\sqrt{5}))=\dfrac{2+\sqrt 5-\frac{1}{2+\sqrt 5}}{2}=\dfrac{(2+\sqrt 5)^2-1}{2(2+\sqrt 5)}=\dfrac{4+2\sqrt 5}{2+\sqrt 5}=2$$, the sought limit is $2$.

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