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let $\{ a_n \}$ be a sequence where for each $n \in \mathbb N$ $ a_n \neq 0 $ and where $\lim_{n \rightarrow \infty} a_n a_{n+1} = L$ with $L \neq 0$

I want to prove that

$\lim_{n \rightarrow \infty} a_n a_{n+3} = L$

and that

$\lim_{n \rightarrow \infty} a_n a_{n+2} \neq -1$

Any ideas?

Thanks!

Edit: Intuitively it's clear but I am looking for a real regorous proof..

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    $\begingroup$ For the first part, note that $$a_{n}a_{n+3}=\frac{(a_{n}a_{n+1})(a_{n+2}a_{n+3})}{a_{n+1}a_{n+2}}.$$Similarly, for the second part, note that $$a_{n}a_{n+2}=\frac{(a_{n}a_{n+1})(a_{n+1}a_{n+2})}{a_{n+1}^2}.$$ Since the numerator converges to the positive number $L^2$ and the denominator is positive, the right-hand side is positive for all large $n$, and so, it cannot converge to a negative number. $\endgroup$ – Sangchul Lee Apr 2 at 10:12
  • $\begingroup$ Interesting question! It does not follow that $\lim a_n a_{n+2} = L$, or even that it exists. $\endgroup$ – GEdgar Apr 2 at 13:03
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First part : Because $a_n \neq 0$ for all $n$, you can write $$a_na_{n+3} = \frac{\left( a_n a_{n+1} \right) \left( a_{n+2} a_{n+3}\right) }{a_{n+1} a_{n+2}}$$

Now it is easy to see that it tends to $$ \frac{L \times L}{L} = L$$

Second part : Similarly you can write $$a_{n+1}^2= \frac{\left( a_n a_{n+1} \right) \left( a_{n+1} a_{n+2}\right)}{a_n a_{n+2} }$$

Suppose that $a_n a_{n+2}$ tends to $-1$ ; then you would deduce that $a_{n+1}^2$ tends to $-L^2$ which is impossible because $-L^2 < 0$ and $a_{n+1}^2 > 0$ for all $n$.

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$a_na_{n+1} \to L$, $a_{n+1}a_{n+2} \to L$,$a_{n+2}a_{n+3} \to L$. Multiply the first and the third and divide by the second to get $a_na_{n+3} \to L$. For the second part note that $a_na_{n+2} a_{n+1}^{2} \to L^{2}$. Can you see why $a_na_{n+2}$ cannot be negative for large $n$?.

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