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Let $a_{n+1}=\dfrac{10}{a_n}-3$, $a_1=10$ then find the limit $\lim\limits_{n \to \infty} a_n$

My Try :

$$a_2=-2 \ \ ,a_3=-8 \ \, a_4=-4.25 \ \ a_n <0$$

thus visthe monotone convergence theorem $$\lim\limits_{n \to \infty}=l$$

so: $$l=\dfrac{10}{l}-3 \to l^2+3l=10 \to l=2 , -5 $$

it is right ?

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  • $\begingroup$ Limit should be $-5$. You would need to show that $a_n$ is bounded (for all $n$) and increasing/decreasing to be able to apply the monotone convergence theorem. $\endgroup$ – The Phenotype Jan 27 '18 at 21:03
  • $\begingroup$ If $a_2=-2$ and $-2$ is a fixed point of the function, then $a_3=-2$, roo. $\endgroup$ – Thomas Andrews Jan 27 '18 at 21:03
  • $\begingroup$ $l=\dfrac{10}{l}-3 \to l^2+3l=10 \to l=\color{red}{2 ,- 5}$ $\endgroup$ – Donald Splutterwit Jan 27 '18 at 21:04
  • $\begingroup$ You should get $a_4=-4.25$ $\endgroup$ – Thomas Andrews Jan 27 '18 at 21:05
  • $\begingroup$ It can't be right if you have two different values. If it converges it can only converge to one. $\endgroup$ – fleablood Feb 3 '18 at 1:46
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When you have a sequence of the form $a_{n+1}=f(a_n)$ that apparently does not lead to a closed formula for $a_n$, then you have to study the function $f(x)$.

When you graph the curve $y=f(x)=\dfrac{10}x-3$ in blue and $y=x$ in red, you notice there are two intersection points.

These are called fixed points of $f$ since $f(x)=x$. Once solved this gives $x=2$ or $x=-5$.

If the sequence would converge to $\ell$, the continuity of $a_{n+1}=f(a_n)$ will lead to $f(\ell)=\ell$ so $\ell$ will be one of the two fixed points.

On the graph we can see that $2$ is a repulsive point, and $-5$ an attractive point.


Since we are not required to do the full study for all initial seeds fo the sequence, but only for $a_1=10$, we will focus on showing it converges to $-5$.

We can see that the convergence is not a staircase (monotonic convergence) but a spiral. This means we have to show that $a_{2n}$ and $a_{2n+1}$ are both monotonic but of opposite direction.

To prove this we have to:

  • study the sign of $a_{n+2}-a_n$, this is equivalent of studying the sign of $f(f(x))-x$.
  • show that $-5$ is squeezed between $a_n$ and $a_{n+1}$, this is equivalent of studying the sign of $f(x)+5$.

First notice that $x<0\implies f(x)<0$ so as soon as $a_{n_0}<0$ then all subsequent $a_n$ with $n\ge n_0$ are also negative.

Since $a_2<0$ we will select $n_0=2$.

$f(x)+5=\dfrac {10}x-3+5=\dfrac{2(x+5)}x\quad\begin{cases} > 0 & x\in]-\infty,-5[\\<0 & x\in]-5,0[\end{cases}$

So if $a_n<-5$ then $a_{n+1}>-5$ and vice-versa and $-5$ is squeezed between $a_n$ and $a_{n+1}$ for $n\ge 2$

$f(f(x))-x=\dfrac{10}{\frac{10}{x-3}}-3-x=\dfrac{3(x+5)(x-2)}{10-3x}\quad\begin{cases} > 0 & x\in]-\infty,-5[\\<0 & x\in]-5,0[\end{cases}$

So $a_{n+2}>a_{n}$ for $a_n<-5$ and $a_{n+2}<a_n$ for $a_n>-5$.

Since $a_2>-5$ then $\begin{cases}a_{2n}>-5 & a_{2n}\searrow\\a_{2n+1}<-5 & a_{2n+1}\nearrow\end{cases}$

Now we can apply monotonicity theorem and $a_{2n}\to -5$ and $a_{2n+1}\to -5$.

This means that $a_n\to -5$.

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Note that once $x$ becomes negative, it remains negative for the rest of time; so we might as well start with $a_3 = -8$ and continue under the fact that $a_n < 0$.

Now, consider $$a_{n+2} = \frac{19 a_n-30}{10-3a_n}$$

Note that this sequence is decreasing in the even terms, and increasing in the odd terms, given the two starting values of $-8$ and $-\frac{17}{4}$: this is because for $x < -5$, we have $-6 + \frac{30+x}{10-3x} < -5$, so inductively the odd-index terms are all less than $-5$; and similarly the even-index terms are all greater than $-5$.

Hence the even terms are decreasing and bounded below; and the odd terms are increasing and bounded above.

Since the two subsequences do converge, it's now legal to do your trick and show that they converge to $-5$ by solving $l = \frac{19l - 30}{10-3l}$.


Note that one can find explicit closed forms for the even and the odd sequences, using generating functions. But that is massive overkill here.

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There is no monotone convergency with this sequence $$a_1=10, a_2=-2, a_3=-8, a_4=-4.25, a_5=-5.35295, a_6=-4.86813$$ is not monotone. It oscillates, so different techniques are required.

Let's look at the function $f(x)=\frac{10}{x}-3$, s.t $a_{n+1}=f(a_n)$ and $f'(x)=-\frac{10}{x^2}$


Preliminary. You have spotted a fixed point $x_0=-5$ which is also an attracting fixed point since $\left|f'(x_0)\right|=\frac{2}{5}<1$. So, there is a vicinity of $x_0$ such that if $a_n$ falls into it from some $n$ onwards, then the sequence will be "attracted" to $x_0$. This is good but not enough.


Technical proof, using Banach fixed-point theorem (BFPT). We see that starting from $n=3$, $a_n\in [-10,-4]$, this is true because

$$\color{red}{x \in [-10,-4]} \Rightarrow -10 \leq x \leq -4 \Rightarrow \\ 10 \geq -x \geq 4 \Rightarrow -\frac{1}{4} \leq \frac{1}{x} \leq -\frac{1}{10} \Rightarrow \\ -10<-\frac{11}{2} \leq \frac{10}{x} -3 \leq -4 \Rightarrow \\ \color{red}{f(x) \in [-10,-4]}$$

And applying MVT $$\forall x<y \in [-10,-4], \exists \xi \in (x,y) : \left|f(x)-f(y)\right|=|f'(\xi )||x-y|\leq\left|\frac{10}{16}\right||x-y|$$

Because $\left|\frac{10}{16}\right|<1$, according to BFPT, there is a limit of the sequence $a_n$ on $[-10,-4]$ (obviously unique), so it's $-5$. More theory here.

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