If $\lim_{n\to\infty}a_na_{n+1}=\lim_{n\to\infty}a_na_{n+2}=1$ and $a_n>0$ find $\lim_{n\to\infty}a_n$

Question

If $\lim_{n\to\infty}a_na_{n+1}=\lim_{n\to\infty}a_na_{n+2}=1$ and $a_n>0$ prove that the limit $\lim_{n\to\infty}a_n$ exists and find it's value.

I tried dividing by $a_n$ since it's positive but didn't get very far. Any tips?

Answer 1

$$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\sqrt{\frac{(a_{n-1}a_n)(a_na_{n+1})}{a_{n-1}a_{n+1}}}=\sqrt{\frac{\lim_{n\to\infty}(a_{n-1}a_n)\cdot\lim_{n\to\infty}(a_na_{n+1})}{\lim_{n\to\infty}(a_{n-1}a_{n+1})}}=\sqrt{\frac{1\cdot 1}{1}}=1$$

which is valid reasoning as all $a_n\gt 0$ and all three limits in the third expression exist (and are equal to $1$).

Answer 2

Divide the two limits you find that $a_n\sim a_{n+1}$ so from the first limit we get that $a_n^2\sim 1$ and since $a_n>0$ then the desired limit is 1.