Recurrent sequence limit

Question

Let $a_n$ be a sequence defined: $a_1=3; a_{n+1}=a_n^2-2$ We must find the limit: $$\lim_{n\to\infty}\frac{a_n}{a_1a_2...a_{n-1}}$$

My attempt

The sequence is increasing and does not have an upper bound. Let $b_n=\frac{a_n}{a_1a_2...a_{n-1}},n\geq2$. This sequence is decreasing(we have $b_{n+1}-b_{n}<0$. How can I find this limit?

Answer 1

since take $a_{1}=x+\dfrac{1}{x},x>1$,then $$a_{2}=x^2+\dfrac{1}{x^2},a_{3}=x^4+\dfrac{1}{x^4}\cdots,a_{n}=x^{2^{n-1}}+\dfrac{1}{x^{2^{n-1}}}$$ so we have $$a_{1}a_{2}\cdots a_{n}=\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}\right)\cdots\left(x^{2^{n-1}}+\dfrac{1}{x^{2^{n-1}}}\right)=\dfrac{x^{2^n}-\dfrac{1}{x^{2^n}}}{x-\dfrac{1}{x}}$$ so $$\dfrac{a_{n+1}}{a_{1}a_{2}\cdots a_{n}}=\left(x-\frac{1}{x}\right)\dfrac{x^{2^n}+\dfrac{1}{x^{2^n}}}{x^{2^n}-\dfrac{1}{x^{2^n}}}\to \dfrac{x^2-1}{x},n\to+\infty$$ and $$\left(x-\dfrac{1}{x}\right)^2+4=\left(x+\dfrac{1}{x}\right)^2=9\Longrightarrow x-\dfrac{1}{x}=\sqrt{5}$$ another approach $$a^2_{n+1}-4=a^4_{n}-4a^2_{n}=a^2_{n}(a^2_{n}-4)=a^2_{n}a^2_{n-1}(a^2_{n-1}-4)=\cdots=a^2_{n}a^2_{n-1}\cdots a^2_{1}(a^2_{1}-4)$$ so we have $$\dfrac{a_{n+1}}{a_{1}a_{2}\cdots a_{n}}=\sqrt{a^2_{1}-2}\cdot \dfrac{a_{n+1}}{\sqrt{a^2_{n+1}-4}}\to\sqrt{a^2_{1}-4}=\sqrt{5},a_{n+1}\to+\infty$$ since $a_{n+1}\to +\infty,n\to +\infty$

Answer 2

Dedicated to @Ant who asked "How did you come up with that?" and to all those that might be interested by a (somewhat) different, possibly more "natural" answer, i.e., @Almagest @David @Eugen Covaci @Turambar.

An alternate change of variable is possible, based on hyperbolic trigonometry. The treatment I am going to give is in fact very parallel to the excellent solution of @Geromty. Setting:

$$a_1=2 \cosh(y)=3$$

Due to relationship $\cosh(2\alpha)=2 \cosh^2(\alpha)-1$, by an immediate recurrence:

$$a_k=2 \cosh(2^{k-1}y)$$

Thus:

$$\Pi_{k=1}^{n}a_{k}=2^n\Pi_{k=0}^{n-1} \cosh(2^ky)=\dfrac{\sinh(2^ny)}{\sinh y} \ \ \ (1)$$

The last closed form formula is a classical result (see Appendix).

Thus $$\dfrac{a_{n+1}}{a_{1}a_{2}\cdots a_{n}}= 2\cosh(2^ny)\dfrac{\sinh(y)}{\sinh(2^ny)}=2\sinh(y) \dfrac{\cosh(2^ny)}{\sinh(2^ny)}$$

which tends (from above), when $n \rightarrow \infty$, to $$2\sinh(y)=2\sqrt{\cosh^2y-1}=2\sqrt{\dfrac{9}{4}-1}=\sqrt{5}.$$

Appendix: Proof of the second equality in (1).

Let us first recall the formula:

$$\sinh(2 \alpha)=2 \sinh(\alpha)\cosh(\alpha) \ \ \ \ (2)$$

Let $P_n$ be the product in (1). Let us multiply $P_n$ by $\sinh(y)$. Then, using (2), one gets $$2^{n-1} \sinh(2y)\cosh(2y)\cosh(4y)\cosh(8y)...\cosh(2^{n-1}y).$$ We can anew apply formula (2), and again in cascade, reducing finally in this way the initial product ($\sinh(y)P_n$) to $\sinh(2^ny)$.

Remark: there is, evidently, a similar result in circular trigonometry $$2^n\Pi_{k=0}^{n-1} \cos(2^ky)=\dfrac{\sin(2^ny)}{\sin y}.$$