# sequence of positive numbers satisfying $a_{n+1}=\frac{2}{a_n+a_{n-1}}$, prove it converges

Assume that $(a_n)$ is a sequence of positive real numbers satisfying $a_{n+1}=\frac{2}{a_n+a_{n-1}}$ for $n=2,3,\dots$.

Prove that $(a_n)$ is convergent and find the limit.

I have no idea how to prove convergence. I tried to prove that $(a_n)$ is bounded and monotone, but both trials failed.

If convergence is proven, the limit is easy: if $g=\lim a_n$, then $g\ge 0$ because all $a_n$'s are positive, the equivalent relation $a_{n+1}(a_n+a_{n-1})=2$ gives $g(g+g)=2$, so finally $g=1$.

• The sequence is clearly not monotone, because it can't have three consecutive terms on the same side of $1$. It must therefore oscillate some way or another. – Arthur Jun 22 '17 at 11:48
• I think you have to specify what the initial seeds $a_0$ and $a_1$ are. – Jose Arnaldo Bebita-Dris Jun 22 '17 at 11:48
• @JoseArnaldoBebitaDris I think that as long as $a_0$ and $a_1$ positive real numbers, the limit is $1$. So no, it's probably not necessary to specify any more than that in order to find the limit. – Arthur Jun 22 '17 at 11:49
• The seeds $a_0,a_1$ are undefined, so they can be any two fixed positive reals. – tong_nor Jun 22 '17 at 11:53

@Michael Rozenberg's result is true. But, there should be proof of boundedness of the sequence. This is in fact easy induction.

Let $a_0, a_1>0$. Find $\alpha>0$ such that $\alpha<\min(a_0, a_1)\leq \max(a_0,a_1)<\frac1{\alpha}$.

Suppose $\alpha<\min(a_{n-1},a_n)\leq \max(a_{n-1},a_n)<\frac1{\alpha}$. Then by the recurrence, we have $$a_{n+1}=\frac2{a_n+a_{n-1}}< \frac 2{2\alpha} = \frac1{\alpha}$$ and $$a_{n+1}=\frac2{a_n+a_{n-1}}> \frac2{2/\alpha}=\alpha.$$

Thus, it follows by induction that $\alpha<a_n<\frac1{\alpha}$ for all $n\geq 0$.

Then as @Michael Rozenberg did, it follows that $\limsup a_n=a>0$, $\liminf a_n=b>0$ satisfy $ab=1$. Along with $b\leq a$, we obtain $b\leq 1\leq a$.

It seems that one application of the recurrence is not enough for proving the convergence. Thus, we try applying the recurrence twice.

$$a_{n+1}=\frac2{a_n+a_{n-1}}=\frac2{\frac2{a_{n-1}+a_{n-2}}+a_{n-1}}.$$

Let $\{a_{n_k}\}$ be a subsequence of $\{a_n\}$ such that $a_{n_k}\rightarrow a$. Then $$a_{n_k}=\frac2{\frac2{a_{n_k-2}+a_{n_k-3}}+a_{n_k-2}}.$$ Take further subsequence $n_{k_l}$ of $n_k$ such that $a_{n_{k_l}-2}$ and $a_{n_{k_l}-3}$ both converge, to the limits $\beta$, $\gamma$ respectively. Then taking $l\rightarrow\infty$ in the recurrence $$a_{n_{k_l}}=\frac2{\frac2{a_{n_{k_l}-2}+a_{n_{k_l}-3}}+a_{n_{k_l}-2}},$$ we obtain $$a=\frac2{\frac2{\beta+\gamma}+\beta}.$$ Now, replacing $\gamma$ by $a$ makes the RHS larger. Thus, $$a\leq \frac2{\frac2{\beta+a}+\beta} .$$ This yields $$a\left(\frac2{\beta+a}+\beta\right) \leq 2,$$ $$a\left(2+ \beta(\beta+a)\right)\leq 2(\beta+a),$$ $$a\beta(\beta+a)\leq 2\beta,$$ $$a(\beta+a)\leq 2.$$ Replacing $\beta$ by $b$, we have $$a(b+a)\leq 2.$$ Since $ab=1$, we have $a^2\leq 1$ which yields $a\leq 1$.

Therefore, $1\leq a \leq 1$, so $a=1$. By $ab=1$, also we have $b=1$. This prove the convergence of $\{a_n\}$ and the limit is $1$.

• It was beautiful! – Michael Rozenberg Jun 23 '17 at 7:40
• @MichaelRozenberg Yours too! It was very helpful. – Sungjin Kim Jun 23 '17 at 7:43
• Actually no need to rely on $ab=1$. You proved $a(b+a)\le2$, in the same way you could prove $b(a+b)\ge2$, so $b^2\ge a^2$, together with $a \ge b$ we get $a=b$. Well done! – dvb Jan 11 '19 at 1:32
• @dvb That's very nice! – Sungjin Kim Jan 11 '19 at 2:12

I believe the following will help.

Let $\overline{\lim}a_n=a$ and $\underline{\lim}a_n=b$.

It's obvious that $a\geq b$.

Now, let $$\lim\limits_{k\rightarrow\infty}{a_{n_k}}=a$$ and $$\lim\limits_{m\rightarrow\infty}{a_{n_m}}=b.$$ We know that $$a_{n_k}=\frac{2}{a_{n_k-1}+a_{n_k-2}}.$$

Consider $\{k_l\}$ such that $$\lim\limits_{l\rightarrow\infty}a_{n_{k_l}-1}=c$$ and $$\lim\limits_{l\rightarrow\infty}a_{n_{k_l}-2}=d.$$ Thus,$$a=\frac{2}{c+d}\leq\frac{2}{b+b}$$ and we obtain $$ab\leq1.$$ By the same way we'll get $$b\geq \frac{2}{2a}$$ and $$ab\geq1.$$ Id est, $ab=1$.

• $ab=1$ and $a\geq b$ proves that $b\leq 1 \leq a$. It is not enough to conclude that $a=b=1$. – Sungjin Kim Jun 22 '17 at 16:24
• @i707107 I fixed my post. Thank you! – Michael Rozenberg Jun 22 '17 at 18:14
• the $k_l$ subsequence is chosen to make only $a_{n_{k_l}-1}$ convergent, and after that we state that $a_{n_{k_l}-2}$ is forced to be convergent (by the recursion formula and the convergence of two others), right? – tong_nor Jun 22 '17 at 19:52
• @tong_nor Not exactly. We know that $a_{n_k-1}+a_{n_k-2}$ is convergent. Thus, there are $\{k_t\}$ and $\{k_s\}$ for which $\{a_{n_{k_t}-1}\}$ and $\{a_{n_{k_s}-2}\}$ are convergent. Now, let $k_l=k_t+k_s$ and we get that all three sequences are convergent. – Michael Rozenberg Jun 22 '17 at 20:04
• how do you know that $a$ is finite and $b > 0$ ? – mercio Jun 22 '17 at 20:15