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 From the link https://www.deltami.edu.pl/2024/09/klub-44/

          Statement of the problem

          "The sequence $\;a_0,\;a_1,\;a_2,\;...\;$ is defined by the formulas

           $a_0=3,\;a_{n+1}=a_n^2-2,\;n=0,\;1,\;2,...\;.$ Calculate the limit

$$\lim_{n \to \infty}\frac{1}{a_n}\prod_{i=0}^{n-1}a_i$$

            or demonstrate that this limit does not exist."

The demonstration is from the same cited place .

          ANSWER : The limit is  $\frac{1}{\sqrt{5}}.$

          Proof : Personal preamble CiP

             A first observation is that $\;a_n>2.$ (It would be too pedantic to prove this by induction; however: $a_0>2\;$ and assuming $a_n>2\;$ we get

 $a_{n+1}=a_n^2-2>2^2-2=2.$)

          We have, because $a_n>2,$

$\;\;a_{n+1}-a_n=a_n^2-2-a_n=(a_n+1)(a_n-2)>0$, so $a_{n+1}>a_n$, meaning the sequence $(a_n)_{n\geqslant 0}$ is strictly increasing.

         Then we have  $a_n\geqslant n+3$ (an extremely poor estimate). (Induction is equally trivial: $a_0\geqslant 3$, and assuming $\;a_n \geqslant n+3$, result

 $a_{n+1}\geqslant (n+3)^2-2=n^2+6n+7\geqslant n+4.$)

However, from the above it follows that

$$\lim_{n \to \infty}a_n=\infty \;. \tag{1}$$

I wrote all this to justify the statement "Jasne, że $a_n \to \infty$". Next, we copy the quoted solution.

<end Preamble>

          We will define the sequence $(b_n)_{n \geqslant 0}$ by

 $b_0=1,\;b_n=a_0\cdots a_{n-1},\;n=1,\;2,\dots$ 

 and we have to investigate the limit of the sequence $\left ( \frac{b_n}{a_n}\right )_{n \geqslant 0}.$

          We see that

$$b_{n+1}=a_n \cdot b_n \tag{2}$$

and we will prove by Induction that 

$$a_n^2-5\cdot b_n^2=4. \tag{3}$$

     Obvious $\;a_0^2-5\cdot b_0^2=3^2-5\cdot 1=4$, and assuming (3) true, result

$$a_{n+1}^2-5\cdot b_{n+1}^2 \overset{def}{\underset{(2)}{=}}(a_n^2-2)^2-5 \cdot (a_n\cdot b_n)^2=\underline{a_n^4}-4a_n^2+4-\underline{5a_n^2b_n^2}=\underline{a_n^2}\cdot (a_n^2-5b_n)-4a_n^2+4 \underset{(3)}{=}a_n^2\cdot 4-4a_n^2+4=4$$

so (3) also occurs for $\;n+1.$ This complete the Induction.

     Now we have 

$$\left (\frac{b_n}{a_n} \right )^2=\frac{b_n^2}{a_n^2}\underset{(3)}{=}\frac{\frac{a_n^2-4}{5}}{a_n^2}=\frac{a_n^2-4}{5a_n^2}=\frac{1}{5}-\frac{4}{5}\cdot \frac{1}{a_n^2}\xrightarrow{(1)}\frac{1}{5}-\frac{4}{5}\cdot 0=\frac{1}{5}.$$

Therefore $\;\frac{b_n}{a_n} \to \frac{1}{\sqrt{5}}.$

$\blacksquare$