Find the limit of the sequence $(a_n)_{n \ge 0}$ given that $a_{n+1}^2=a_na_{n-1}$ and $a_0=2,$ $a_1=16$.

Question

I have the following sequence:

$$a_{n+1}^2=a_na_{n-1}, \forall n \ge 1$$

With $a_0=2$ and $a_1 = 16$. I am also told that the sequence has positive terms. I have to find the limit of the sequence $(a_n)_{n \ge 0}$.

I am not that great at limits/sequences so I'm not sure how should I approach this.

If you calculate a few terms, you'll see it appears to be approaching $8$ — J. W. Tanner, Nov 10 '19 at 16:41
It might be easier to instead think about the sequence $b_n = \log_2a_n$. Then you've got $b_0 = 1$, $b_1 = 4$, and $b_{n + 1} = \frac{1}{2}(b_n + b_{n - 1})$. — Jim, Nov 10 '19 at 16:42
WA says this here $$a(n)\to 8 e^{\frac{1}{3} (-1)^n 2^{1-n} (\log (2)-\log (16))}$$ — Dr. Sonnhard Graubner, Nov 10 '19 at 16:45

score 2 · Accepted Answer · answered Nov 10 '19 at 16:43

2

Hint

Let $b_n=\log(a_n)$ which makes $$2 b_{n+1}=b_n+b_{n-1}$$ which looks to be simple.

answered Nov 10 '19 at 16:43

Claude Leibovici

260,315

Calvin Lin · Answer 2 · 2019-11-10T20:38:21.863

0

Hint: Do the AM version first.

If $b_{n+1} = \frac{b_n + b_{n-1} } { 2}$, then $\lim b_n = \frac{ b_1 + 2 b_0} { 2}$.

Apply to $a_n = 2^{b_n}$.

edited Nov 10 '19 at 20:38

answered Nov 10 '19 at 16:41

Calvin Lin

68,864

When you wrote $b_0+2b_1$, did you mean $2b_0+b_1$? – J. W. Tanner Nov 10 '19 at 16:47
@J.W.Tanner . Agreed. Must be a typo. – DanielWainfleet Nov 10 '19 at 17:48
Thanks, fixed.. – Calvin Lin Nov 10 '19 at 20:38

score 0 · Answer 3 · answered Nov 10 '19 at 18:01

With $a_n=2^{b_n}$, the sequence $b_0,b_1,b_2,b_3...$ is $1, \,3+1,\, 3-1/2,\, 3+1/4, \,3-1/8,...$ Prove that $A(n)\implies A(n+1)$ when $n>0,$ where $A(n)$ is $b_n=3\pm 2^{-(n-1)} \land b_{n+1}=3\mp 2^{-n}$. And so by Induction on $n$ , obtain $b_n\to 3.$

Find the limit of the sequence $(a_n)_{n \ge 0}$ given that $a_{n+1}^2=a_na_{n-1}$ and $a_0=2,$ $a_1=16$.

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