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I was asked to find the limit of:

$a_{n+2}=\sqrt{a_na_{n+1}}$

$a_1=1, a_2=2$

It seems as if the sequence is constant from n=4 and it's value is $a_n=\sqrt{2\sqrt{2}} -\forall{n>3}$

I'd just like to double-check I did thing right.

Thanks!

Community
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Paz
  • 789

3 Answers3

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The sequence is not constant, it is: $$1, 2, 2^\frac{1}{2}, 2^\frac{3}{4}, 2^\frac{5}{8}, 2^\frac{11}{16}, \ldots$$ Hint: If you want to know what it converges to take a look at the sequence of exponents: $$0, 1, \frac{1}{2}, \frac{3}{4}, \frac{5}{8}, \frac{11}{16}, \ldots$$ and figure out what it's recursive rule is and what it converges to.

Jim
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Take logs of both sides and define $b_n = \log{a_n}$. Then

$$2 b_n - b_{n-1}-b_{n-2} = 0$$ $$b_0=0$$ $$b_1=\log{2}$$

This is a constant coefficient difference equation with solution $b_n=A r^n$,where $r$ satisfies

$$2 r^2-r-1=0$$

with solutions $r_+ = 1$ and $r_-=-1/2$. Thus

$$b_n = A + B \left (-\frac{1}{2} \right )^n$$

with

$$A+B=0$$ $$A-\frac{1}{2} B = \log{2}$$

The $$b_n=\frac{2}{3} \log{2} \left [ 1 + \left (-\frac{1}{2} \right )^{n+1} \right]$$

Then going back to the original sequence:

$$a_n = 2^{\frac{2}{3}\left [ 1 + \left (-\frac{1}{2} \right )^{n+1} \right]}$$

$$\lim_{n \rightarrow \infty} a_n = 2^{2/3}$$

Ron Gordon
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Using the trick described in answers to this post, it becomes quite simple.

We have the recurrence relation $a_{n+2}=\sqrt{a_n\cdot a_{n+1}}$.

Multiplying both sides by $\sqrt{a_{n+1}}$, $\quad a_{n+2}\cdot\color{green}{\sqrt{a_{n+1}}}=\sqrt{a_n\cdot a_{n+1}\cdot\color{green}{a_{n+1}}}\tag*{}$

or, $a_{n+2}\cdot\sqrt{a_{n+1}} = a_{n+1}\cdot\sqrt{a_n} \tag{$\star$}$

The relation $(\star)$ is true for all $n\in\mathbb Z^+$. Using it recursively, we get:

$\begin{align}a_{n+2}\cdot\sqrt{a_{n+1}} &= a_{n+1}\cdot\sqrt{a_n} \\ &=a_{n}\cdot \sqrt{a_{n-1}}\\ &=a_{n-1}\sqrt{a_{n-2}}\\&\quad\quad\vdots\\&=a_2\cdot\sqrt{a_1}\end{align}\tag*{}$

Taking limit $n\to\infty$, we get: $$\lim_{n\to\infty}a_{n+1}\cdot\sqrt{a_n}=a_2\cdot\sqrt{a_1}$$

If we let $\lim a_n=L$, we have:

$ L\sqrt L=a_2\sqrt{a_1}\quad$ or, $\quad L=(a_2\sqrt{a_1})^{2/3}$.

Taking $a_1=1$ and $a_2=2$, the limit is $L=2^{2/3}=\sqrt[3]{4}$.

Nothing special
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  • How do you know that $\lim a_n$ exists? – Gary Feb 10 '24 at 22:43
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    @Gary I'm naive to rigorous math and didn't investigate this until you asked. The sequence is bounded in $[1,2]$ by the property of geometric mean. The odd subsequence is increasing while the even subsequence is decreasing. By monotone convergence theorem, both the subsequences converge. And they converge to the same point; because (in accordance with the recurrence relation) we need $L_e=\sqrt{L_eL_o}$ where $L_e=\lim a_{2n}$ and $L_o=\lim a_{2n-1}$. – Nothing special Feb 10 '24 at 22:51