How I can find a limit for this recursively defined sequence?
$$a_0>0, a_{n+1}=\frac{a_{n}+2}{3a_{n}+2}$$
I'm particularly interested in answers involving concepts like contractive sequences and fixed points.
Many thanks.
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There are two fixed points, $\frac23$ and $-1$. Let's look at the stability near each of these. Let $$ f(x)=\frac{x+2}{3x+2}\tag{1} $$ Then $$ f'(x)=-\frac43\frac1{(3x+2)^2}\tag{2} $$ Since $f'\left(\frac23\right)=-\frac13$ and $f'(-1)=-\frac43$, $\frac23$ is a stable fixed point, $|f'(x)|\lt1$, and $-1$ is an unstable fixed point, $|f'(x)|>1$. Let's investigate the stable fixed point. The recursive definition centered on $\frac23$ becomes $$ \left(a_n-\tfrac23\right)=-\frac{\left(a_{n-1}-\tfrac23\right)}{3\left(a_{n-1}-\tfrac23\right)+4}\tag{3} $$ Note that if $a_{n-1}\gt0$, then $3\left(a_{n-1}-\tfrac23\right)+4\gt2$. Thus, $(3)$ implies that $$ \left|a_n-\tfrac23\right|\lt\tfrac12\left|a_{n-1}-\tfrac23\right|\tag{4} $$ $(4)$ guarantees convergence. |
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Use the recurrence relation $$ a_{n+1} - a_{n} = \frac{a_n +2 }{ 3 a_n +2} - \frac{a_{n-1} +2}{3 a_{n-1} +2 } = \frac{4(a_{n-1} -a_{n})}{(3 a_{n-1} +2)(3 a_n +2)}. $$ Since $$ 3 a_n = 3 \frac{a_{n-1} +2 }{3 a_{n-1} +2} > 1 + \frac{1}{a_{n-1}+1} >1, \quad\forall n\geq 1, $$ it follows that $$ |a_{n+1} - a_{n}| < \frac{4}{9} |a_{n} - a_{n-1}|. $$ Iteration gives $$ |a_{n+1} - a_{n}| < \left(\frac{4}{9}\right)^n |a_{1} - a_{0}|. $$ The series $\sum_{n=1}^\infty (a_{n+1}-a_n)$, of positive terms, is dominated by the convergent series $|a_1-a_0| \sum_{n=1}^\infty (4/9)^n$ and so converges. We have $\sum_{n=1}^\infty (a_{n+1}-a_n)= \lim_{n\to \infty} a_n - a_1$ which shows that the limit exists. Then, to find to fixed points we can pass to the limit in the recurrence relation $$ a_{\infty}= \frac{a_{\infty}+2}{3 a_{\infty} +2}, $$ which leads to $a_{\infty}= 2/3.$ |
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A little playing shows that the limit "should" be $\frac{2}{3}$. So it is natural to compute $a_{n+1}-\frac{2}{3}$. We get $$a_{n+1}-\frac{2}{3}=\frac{a_n+2}{3a_n+2}-\frac{2}{3}=\frac{\frac{2}{3}-a_n}{3a_n+2}.$$ Thus $$\left|a_{n+1}-\frac{2}{3}\right|=\left|a_n-\frac{2}{3}\right|\frac{1}{3a_n+2}.$$ In particular, $$\left|a_{n+1}-\frac{2}{3}\right|\lt \frac{1}{2}\left|a_n-\frac{2}{3}\right|.$$ So with each iteration our distance from $\frac{2}{3}$ shrinks by a factor of at least $\frac{1}{2}$. It follows that $\lim_{n\to\infty}a_n=\frac{2}{3}$. |
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$1$. First show that $a_n \in (0,1)$ using induction. $2$. Now if $a_0 < \dfrac23$, using induction, show that $a_n$ is a monotone increasing sequence bounded above by $\dfrac23$. $3$. $\vert \vert \vert^{ly}$, if $a_0 > \dfrac23$, using induction, show that $a_n$ is a monotone decreasing sequence bounded below by $\dfrac23$. $4$. Now recall the completeness of $\mathbb{R}$/ monotone sequence theorem, to conclude that the limit exists. $5$. Now make use of limit laws to show that if $\lim_{n \to \infty} a_n = L$, then $$L = \dfrac{L+2}{3L+2}$$ $6$. Solve the quadratic to get that $L = \dfrac23$. |
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Fix $c\in(0,3^{-1})$ and show that the map $$ \varphi_c:[c,+\infty)\to[c,+\infty):x\mapsto\frac{x+2}{3x+2} $$ is well defined and what is more $$ |\varphi_c(x)-\varphi_c(y)|\leq\frac{4}{(3c+2)^2}|x-y| $$ Hence you can apply Banach fixed point theorem to show that $\varphi_c$ have unique fixed point on $[c,+\infty)$ and this is point is $2/3$. Since this fixed point is the same for all $c$ and $f|_{[c,+\infty)}=\varphi_c$, then $f$ has unique fixed point on $(0,+\infty)$ |
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