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 ?

  • $\begingroup$ 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 $a_n\to -5$.


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.


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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