# Given $\lim_{n \rightarrow \infty} a_n a_{n+1} = L$, how to show: $\lim_{n \rightarrow \infty} a_n a_{n+3} = L$

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..

• 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. Apr 2, 2019 at 10:12
• Interesting question! It does not follow that $\lim a_n a_{n+2} = L$, or even that it exists. Apr 2, 2019 at 13:03

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 $$n$$.
$$a_na_{n+1} \to L$$, $$a_{n+1}a_{n+2} \to L$$,$$a_{n+2}a_{n+3} \to L$$. 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 $$n$$?.