Sequence $x_{n+1} = \sqrt{x_n x_{n-1}}$

Question

Let $0<a<b<\infty$ be real numbers and define the sequence $x_1=a$, $x_2=b$, $x_{n+1} = \sqrt{x_n x_{n-1}}$. Find the limit of $(x_n)_{n=1}^\infty$.

So I have proved that the sequence is a contraction and that $\lim_{n\rightarrow \infty} (x_{2n+2}-x_{2n+1})=0$. Therefore the sequence converges to some number $x$ in the interval $(a,b)$. Numerically, I have found that the limit is actually $x=(ab^2)^{1/3}$, but I'm stuck in actually proving this result analytically. A hint will be the selected answer.

You may suppose that $x_n = a^{q(n)} b^{r(n)}$ and try to find $q(n)$ and $r(n)$. Then limit will be easy-founded — openspace, Aug 11 '20 at 22:05
I'm trying the subsequence $x_1,x_3,x_5,\dots$ which is monotone increasing but I can't see a nice recursion. — user2820579, Aug 11 '20 at 22:08

score 2 · Answer 1 · answered Aug 11 '20 at 22:13

2

Hint : take the $ln$ (after verifying that $ln(x_n)$ is well-defined !)

answered Aug 11 '20 at 22:13

Thanks for your answer. I think it was similar to what flawr suggested (though I cannot tell from the short answer given), but he developed a little bit more the idea. – user2820579 Aug 12 '20 at 18:37

flawr · Accepted Answer · 2020-08-12T07:36:31.653

2

Let's define $y_n = \log(x_n)$. Then we get the linear recurrence

$$y_{n+1} = \frac{y_n + y_{n-1}}{2},$$

which we can immediately solve using standard methods: Therefore $$y_n = u \left(\frac{-1}{2}\right)^n + v$$ for some $u$ and $v$. And therefore $\lim x_n = v$. Using the initial conditions we get $v = \frac{1}{3} (\log(a) + 2\log(b))$. Since $\log$ is continuous we therefore get

$$\lim x_n = \exp(\lim y_n) = \exp(v) = (ab^2)^{1/3}$$

as you already claimed.

edited Aug 12 '20 at 07:36

answered Aug 11 '20 at 22:28

flawr

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Shouldn't be $y_n = u(-1/2)^n\dots$? – user2820579 Aug 11 '20 at 23:14
@user2820579 yes indeed, thakns for pointing it out! – flawr Aug 12 '20 at 07:36

Sequence $x_{n+1} = \sqrt{x_n x_{n-1}}$

2 Answers2