If $a_n>0$ for all $n$, and $\lim_{n\to\infty}a_na_{n+1}=A$, $\lim_{n\to\infty}a_na_{n+2}=B$, $\lim_{n \to \infty}a_na_{n+3}=C$, dis/prove $A=B=C$

Question

I have hard time proving the following:

Let $a_n$ be a sequence such that $a_n>0$ for all $n$ and:

$$\lim_{n \to \infty}a_na_{n+1}=A$$ $$\lim_{n \to \infty}a_na_{n+2}=B$$ $$\lim_{n \to \infty}a_na_{n+3}=C$$

I try to prove or disprove that $A=B=C$.

So far I have managed to prove that $A=B$:

$$A^2=\lim\limits_{n \to \infty} a_n a_{n+1}\cdot\lim\limits_{n \to \infty}a_{n+2} a_{n+3} = \lim\limits_{n \to \infty}\left(a_n a_{n+1} a_{n+2}a_{n+3}\right)$$ $$B^2= \lim\limits_{n \to \infty} a_n a_{n+2}\cdot\lim\limits_{n \to \infty}a_{n+1}a_{n+3} = \lim\limits_{n \to \infty}\left(a_n a_{n+1} a_{n+2}a_{n+3}\right)$$

So we get that $A^2=B^2$ and because $A,B$ are not negative we get that $A=B$.

Answer 1

Hint. Similarly, calculate $C^3$.

Answer 2

You can adapt your prior argument to this new case, with a slightly more complicated product.

Observe that: $$ \begin{align} A^3 &= \prod_{i = 0}^2\lim_{n \to \infty}\left(\alpha_{n + 2i} \cdot \alpha_{n + 1 + 2i}\right)\\ &= \lim_{n \to \infty}\prod_{i = 0}^5\alpha_{n + i}\\ &= \prod_{i = 0}^2 \lim_{n \to \infty}\alpha_{n + i} \cdot \alpha_{n + 3 + i}\\ &= C^3 \end{align} $$ As both $A$, $C$ are nonegative, taking cube roots yields $A = C$ as required.