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Evaluate the value of $$\lim_{n \to \infty}\frac{x_n}{\prod_{i=1}^{n-1} x_i}$$ where $x_n=x_{n-1}^2-2, (x_1=5).$

I've found that $x_n$ is an increasing sequence so that the limit becomes an indeterminate form of $(\frac{\infty}{\infty})$. I can't think of any other workable ideas though. Some hints are appreciated.

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  • $\begingroup$ What values did you get for the first four values of the quotient? And what do you get if you square these? $\endgroup$
    – Henry
    Feb 26, 2020 at 16:27
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    $\begingroup$ Write $x_1=a+\frac{1}{a}$ for some $a> 1$. You should be able to see that $$x_n=a^{2^{n-1}}+\frac{1}{a^{2^{n-1}}}.$$ Moreover, $$\left(a-\frac1a\right)x_1x_2\cdots x_{n-1}=a^{2^{n-1}}-\frac{1}{a^{2^{n-1}}}.$$ $\endgroup$
    – Batominovski
    Feb 26, 2020 at 16:30
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    $\begingroup$ According to Mathematica, the solution to the recurrence is $x_n=2\cos\left(2^{n-1}\cos^{-1}\left(\frac52\right)\right)$. Though I doubt the point of this exercise is to solve it exactly... $\endgroup$
    – user170231
    Feb 26, 2020 at 16:31
  • $\begingroup$ Does the use of inverse cosine of a number not in $[-1,1]$ not require further explanation? This formula raises more questions than it answers, the argument of $\cos$ must be complex, as from the problem clearly $x_n\to\infty$ $\endgroup$
    – vujazzman
    Feb 26, 2020 at 17:24
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    $\begingroup$ @vujazzman You can extend $\cos$ to the whole complex plane, so $\cos^{-1}x$ is defined for real numbers $x$ s.t. $|x|>1$. When $x$ is a real number s.t. $|x|>1$, $\cos^{-1}x$ is just $i\cosh^{-1}x$ (well, this is the principal value). So the formula can be instead written as $$x_n=2\cosh\Biggl(2^{n-1}\cosh^{-1}\left(\frac52\right)\Biggr).$$ This version probably makes sense to more people. $\endgroup$
    – Batominovski
    Feb 26, 2020 at 17:28

1 Answer 1

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As you noticed $x_n \to \infty$. We have that:

$$x_n=x_{n-1}^2-2\Rightarrow x_n^2=x_{n-1}^4-4x_{n-1}^2+4\Rightarrow x_n^2-4=x_{n-1}^2(x_{n-1}^2-4)$$

Therefore:

$$x_n^2-4=x_{n-1}^2(x_{n-1}^2-4)=x_{n-1}^2x_{n-2}^2(x_{n-2}^2-4)=\ldots=(x_1^2-4)\prod_{i=1}^{n-1}x_i^2$$

Thus:

$$\prod_{i=1}^{n-1}x_i=\frac{\sqrt{x_n^2-4}}{\sqrt{x_1^2-4}}$$

Therefore:

$$\lim_{n\to \infty} \frac{x_n}{\prod_{i=1}^{n-1}x_i}=\lim_{n\to \infty}\frac{x_n\sqrt{x_1^2-4}}{\sqrt{x_n^2-4}}=\lim_{n\to \infty}\frac{\sqrt{x_1^2-4}}{\sqrt{1-\frac{4}{x_n^2}}}=\sqrt{x_1^2-4}$$

In the case of this question, the answer is $\sqrt{21}$.

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    $\begingroup$ +1. It may be worth noting a little more explicitly that this is saying the product is $\frac{\sqrt{x_n^2-4}}{\sqrt{x_1^2-4}}$ and the limit of the quotient is $\sqrt{x_1^2-4}$, which is $\sqrt{21}$ when $x_1=5$ $\endgroup$
    – Henry
    Feb 26, 2020 at 16:33
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    $\begingroup$ @Henry, edited accordingly. $\endgroup$
    – LHF
    Feb 26, 2020 at 16:36
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    $\begingroup$ Execution of skillful mathematical acumen!! I enjoyed your solution. $\endgroup$
    – Lawrence Mano
    Feb 26, 2020 at 16:51

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