Compute $\lim_\limits{n\to\infty}a_n$ where $a_{n+2}=\sqrt{a_n.a_{n+1}}$

I managed to show that the limit exists, but I don't know how to compute it.

EDIT:

There are initial terms: $a_1=1$ and $a_2=2$.

Solution 1:

Note that $$ a_{n+2}\sqrt{a_{n+1}}=a_{n+1}\sqrt{a_n} =\cdots =a_2\sqrt{a_1}=2$$

Hence limit is $2^\frac{2}{3}$.

Solution 2:

With $b_n = \log a_n$ (as suggested in a comment) the iteration formula becomes $$ b_{n+2} = \frac 12 (b_n + b_{n+1}) \, . $$

Computing the first iterates numerically leads to the conjecture that the sequence converges to $(b_0 + 2b_1)/3\, $.

Therefore we define $$ b := \frac{b_0 + 2b_1}3 \, , \quad c := \frac{b_1 - b_0}3 \, . $$ Then $$ b_0 = b - 2c \,, \quad b_1 = b + c \\ b_2 = b - \frac c2\,, \quad b_3 = b + \frac c4 \\ b_4 = b - \frac c8\,, \quad b_5 = b + \frac c{16} \\ $$ and generally $$ b_n = b + \frac{(-1)^{n-1}c}{2^{n-1}} $$ It follows that $$ \lim_{n \to \infty} b_n = b = \frac{b_0 + 2b_1}3 $$ and therefore $$ \lim_{n \to \infty} a_n = \sqrt[3]{a_0 a_1^2} $$ In your case ($a_0 = 1$, $a_1 = 2$) the limit is $2^{2/3} \approx 1.587401$.

Solution 3:

Looking at the sequence, we first see that if $a_n' = \alpha a_n$ and $a_{n+1}' = \alpha a_{n+1}$, then also $a_{n+2}' = \alpha a_{n+2}$, and therefore also for the limit $a := \lim_{n\to\infty}a_n$ we have $a' = \alpha a$. Therefore we can write $$a = a(a_1, a_2) = a_1 a(1, a_2/a_1) =: a_1 f(a_2/a_1).$$ Now what properties does $f$ have? Well, we can of course start the sequence at every position, therefore we have $$f(x) = a(1, x) = a(x, \sqrt{x}) = x a(1, x^{-1/2}) = xf(x^{-1/2}).$$ Also we can see that if $a_n' = a_n^\alpha$ and $a_{n+1}' = a_{n+1}^{\alpha}$, then also $a_{n+2}' = a_{n+2}^\alpha$, and thus $$f(x^\alpha) = f(x)^\alpha.$$ Thus we have $$f(x) = xf(x^{-1/2}) = x(f(x))^{-1/2}$$ and therefore $$f(x)^{3/2} = x \implies f(x) = x^{2/3}$$ Thus we have $$a = a_1 f(a_2/a_1) = a_1 \left(\frac{a_2}{a_1}\right)^{\frac23} = a_1^{1/3} a_2^{2/3}.$$ In particular, with $a_1=1$ and $a_2=2$ we get $$\lim_{n\to\infty} a_n = 1^{1/3}2^{2/3} = \sqrt[3]{4}.$$

Solution 4:

We have $a_2\gt a_1\gt0$ and $a_{n+2}^2=a_{n+1}a_n$.

So,$$\begin{align} &a_3^2=a_2a_1\\ \implies&a_3^2\gt a_1^2\\ \implies &a_3\gt a_1 \end{align}$$

and $$\begin{align} &a_3^2=a_2a_1\\ \implies &a_3^2\lt a_2^2\\ \implies &a_3\lt a_2 \end{align}$$

So, we have $$\color{red}{a_2\gt a_3\gt a_1\gt 0}$$

Similarly, $$\begin{align} &a_4^2=a_3a_2\\ \implies &a_4^2\lt a_2^2\\ \implies &a_4\lt a_2 \end{align}$$ and $$\begin{align} &a_4^2=a_3a_2\\ \implies &a_4^2\gt a_3^2\\ \implies &a_4\gt a_3 \end{align}$$

So, we have $$\color{red}{a_2\gt a_4\gt a_3\gt a_1\gt 0}$$

Now, following this pattern, one can easily see that this sequence is bounded below by $0$ and bounded above by $a_2$.

Now, consider two subsequences $\color{blue}{v_n=a_{2n-1}}$ and $\color{blue}{w_n=a_{2n}}$.

one can easily see that $v_n$ is increasing subsequence bounded above by $a_2$, since, $\color{blue}{0\lt a_1\lt a_3\lt \dots \lt a_2}$, and $w_n$ is decreasing subsequebce and bounded below by $0$, since, $\color{blue}{a_2\gt a_4\gt \dots \gt 0}$.

So, they must converge. Suppose $\color{blue}{\lim v_n=l_1}$ and $\color{blue}{\lim w_n=l_2}$.

Now, we have, $$\begin{align} &a_{n+2}^2=a_{n+1}a_n\\ \implies &\lim a_{n+2}^2=\lim a_{n+1}\times\lim a_n\\ \implies &l_2^2=l_1\times l_2\\ \implies &\color{blue}{l_2=l_1=l(\text{say})} \end{align}$$

here, without loss of generality, we assumed $n$ is even. So, the mother sequence $\{a_n\}$ converges to $l$ i.e. $\color{blue}{\lim a_n=l}$

Now write $$\begin{align} \color{green}{\cancel{a_3^2}}&=a_2\times a_1\\ \color{purple}{\cancel{a_4^2}}&=\color{green}{\cancel{a_3}}\times a_2\\ \color{orange}{\cancel{a_5^2}}&=\color{purple}{\cancel{a_4}}\times \color{green}{\require{cancel}\cancel{a_3}}\\ &\vdots\\ \color{navy}{a_{n+1}^\cancel{2}}&=\color{maroon}{\cancel{a_{n}}}\times \color{teal}{\cancel{a_{n-1}}}\\ a_{n+2}^2&=\color{navy}{\cancel{a_{n+1}}}\times \color{maroon}{\cancel{a_n}}. \end{align}$$

Now, multiplying all the terms, we have $$\begin{align} &a_{n+2}^2 \times a_{n+1}=a_2^2 a_1\\ \implies &\lim a_{n+2}^2 \times \lim a_{n+1}=a_2^2 a_1\\ \implies &l^2\times l= a_2^2 a_1\\ \implies &l^3=a_2^2 a_1\\ \implies &\color{red}{l=\sqrt[3]{a_2^2 a_1}}. \end{align}$$

And, here the particular case, $a_1=1$ and $a_2=2$, so, $$\lim_{n\to\infty} a_n = 1^{1/3}2^{2/3} = \sqrt[3]{4}.$$