limit of : $a_{n+2} =\frac{1}{a_n} + \frac{1}{a_{n+1}}$

Question

$a_n$ is a real sequence, $a_1,a_2$ are positive and for all $n>2$ : $$ a_{n+2} =\frac{1}{a_{n+1}} + \frac{1}{a_{n}}.$$ Prove that: $\displaystyle \lim_{n\to \infty} a_n$ exists, then find it.

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Mathematica gave me $\sqrt{2}$ as an approximate limit, I tried to eliminate $a_{n+2}$ then do some work with Stolz lemma but I failed, Are there any strategy to find the asymptotic expansion for this kind of sequences (I can do it only if it was a first order) ?

Any help is appreciated (if you find it too easy, just post hints).

Answer 1

It is easy to show that the sequence is bounded. If $\sqrt{2}/M \le a_n,a_{n+1} \le \sqrt{2}M$, then also $\sqrt{2}/M \le a_{n+2} \le \sqrt{2}M$.

In fact, if we let $b_n=\max(|\log (a_n/\sqrt{2})|,|\log (a_{n+1}/\sqrt{2})|)$, then $b_{n+1}\le b_n$. Therefore (since $b_n$ is non-negative) $b_n$ converges to its infimum. Let us denote $\lim_{n\to\infty} b_n = B$. All that is left is to do is prove that $B=0$. Suppose that $B>0$, then for some $N$, we have $B-\epsilon<b_n<B+\epsilon$ for all $n>N$. So $|\log (a_{N}/\sqrt{2})|<B+\epsilon$ and $|\log (a_{N+1}/\sqrt{2})|<B+\epsilon$. If both $a_N$ and $a_{N+1}$ are on different sides of $\sqrt{2}$, then $|\log (a_{N+2}/\sqrt{2})|<B-2\epsilon$ (with $C$ some constant) and the same for $a_{N+3}$. If both are on the same side then $a_{N+2}$ is on the other side, and the argument is the same. In any case we have that $b_{N+3}<B-2\epsilon$ and so a contradiction.

Answer 2

If $a_n$ converges towards $a$, then the limit satisfies $a = \frac 1 a + \frac 1 a$.

Answer 3

Maybe prove this. Suppose $$ {\sqrt{2}}(1+\varepsilon)>a_{n+1}>\frac{\sqrt{2}}{1+\varepsilon},\qquad {\sqrt{2}}(1+\varepsilon)>a_{n}>\frac{\sqrt{2}}{1+\varepsilon}. $$ If $a_{n+1},a_n$ are on the same side of $\sqrt{2}$, then $a_{n+2}$ is on the opposite side and $$ {\sqrt{2}}(1+\varepsilon)>a_{n+2}>\frac{\sqrt{2}}{1+\varepsilon}. $$ If $a_{n+1},a_n$ are opposite sides of $\sqrt{2}$, then $$ {\sqrt{2}}\left(1+\frac{\varepsilon}{2}\right)>a_{n+2}>\frac{\sqrt{2}}{\displaystyle 1+\frac{\varepsilon}{2}}. $$

Answer 4

Let's rewrite the equality as follows:

$$a_{n+1} =\frac{1}{a_{n}} + \frac{1}{a_{n-1}}.$$ Now, let's subtract $a_n$ from both sides of the equation:

$$\frac{a_{n+1}-a_n}{n+1-n} =\frac{1}{a_{n}} + \frac{1}{a_{n-1}}-a_n$$ Now, as n goes to infinity lets denote

$$a_n=a$$ Then approximately:

$$\frac{da}{dn}=\frac{2}{a}-a$$ or

$$\frac{a\;da}{2-a^2}=dn$$ After integrating we get:

$$\ln(2-a^2)=-2n+\text{const}$$ or

$$a^2=2-\text{const}\;e^{-2n}$$ A conclusion: $a_n$ is approaching the limit value $\sqrt{2}$ exponentially.

Answer 5

It isn't a proof (at all), but rarely do I miss a chance to use improper fractions. We observe that, $a_n \neq 0$ for all $n \in \mathbb{N}$. (This does not mean that the limit is non-zero, or even exists.) $$ a_{n+2} = \frac{1}{a_{n+1}}+\frac{1}{a_n} = \frac{1}{\frac{1}{a_n+a_{n-1}}} + \frac{1}{\frac{1}{a_{n-1}+a_{n-2}}} $$

So in the limit this becomes a never ending stack, $$ \lim_{n\to\infty} a_n = \frac{1}{\frac{1}{\frac{1}{\vdots}+\frac{1}{\vdots}}+\frac{1}{\frac{1}{\vdots}+\frac{1}{\vdots}}} + \frac{1}{\frac{1}{\frac{1}{\vdots}+\frac{1}{\vdots}}+\frac{1}{\frac{1}{\vdots}+\frac{1}{\vdots}}} $$

$$ \lim_{n\to\infty}a_n = \frac{1}{\lim_{n\to\infty}a_n} + \frac{1}{\lim_{n\to\infty}a_n} = \frac{2}{\lim_{n\to\infty}a_n} $$

Finally, we have; $$ (\lim_{n\to\infty}a_n)^2 = 2 \implies \lim_{n\to\infty}a_n = \sqrt{2} $$

Sorry that this isn't a rigorous proof, but it sure is a fun way to look at the problem.