Proving the convergence of the sequence defined by $x_1=3$ and $x_{n+1}=\frac{1}{4-x_n}$

Question

Consider the sequence defined by $$x_1=3 \quad\text{and}\quad x_{n+1}=\dfrac{1}{4-x_n}$$

I can calculate limit by assuming limit exist and solving quadratic equation, but I first wanted to give existence of limit.

I tried to show given sequence is decreasing and bounded below by 0.

I used derivative test as $$f^\prime(x)=\frac{1}{(4-x)^2}$$ but form this I am not able to show

Also, I tried to shoe $x_{n+1}-x_n<0$ but that way also I am not succeed.

Please tell me how to approach such problem

Answer 1

You will want to show $x>f(x)$ in order to establish that the sequence is decreasing. So consider $g(x) = x - \dfrac{1}{4-x}.$

Since $g'(x) = 1-\dfrac{1}{(4-x)^2} =\dfrac{(x-5)(x-3)}{(4-x)^2},$ it suggest that your intuition true if $x_n\leq 3$ for $n>1$, which can be easily established by induction.

Answer 2

To show that $x_n$ is decreasing, use induction to prove that $x_n> x_{n+1}$. To show that $x_n$ is bounded below by zero show a stronger assertion that $0<x_n<2+\sqrt{3}$ by using induction.

Answer 3

Suppose $p(y)=y^2-4y+1=0$ and let $x_n=y+z_n$ then $$y+z_{n+1}=\frac 1{4-y-z_n}$$ so that $$-yz_n+(4-y)z_{n+1}=y^2-4y+1=0$$ and $$z_{n+1}=\frac y{4-y}z_n$$

So if $0\lt y\lt 2$ the error term $z_n$ has the same sign and is decreasing in magnitude in constant proportion so tends to zero. It is easy to test that there is a root of the quadratic in the required range by noting $p(0)=1, p(2)=-3$

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Note: the equations hold for the other possible value of $y$ too, but the inequality does not then apply to prove that the error term tends to zero. Also this was done without computing $y$.

Answer 4

It can be approached in a graphical manner:

• Draw the graph of $y = \frac{1}{4-x}$ to scale while marking the essentials.
• Asymptote at $x=4$; Value at $x = 3$ is $1$.
• Comparing it to the previous value of the sequence would require the plot of $y=x$ on the same axes.
• Mark the intersection as $x=2-\sqrt3$ whereas $x=2+\sqrt3$ is near $x=4$.

If through, notice that starting the sequence from $x=3$ means that the next value is $1$ from the hyperbola which is well below the straight line. Now to get the next value put $x=1$ and get the next value from the hyperbola, which is again less than $1$ as the straight line depicts.

If you follow the pattern, you would tend to reach the intersection $x=2-\sqrt3$ as the gap between both the curves decreases to zero which gives the limit of the sequence as $x=2-\sqrt3$ (the limit only, not one of the terms of the sequence, since these are all rational numbers).

Also, one can thus say that if $x_1 \in (2-\sqrt3,2+\sqrt3)$ then the sequence would be decreasing and would converge at $x=2-\sqrt3$ and that all the terms $x \in (2-\sqrt3,2+\sqrt3)$.

Answer 5

Hint: after finding the limit $l$ from $l={1\over 4-l}$ to make sure that the sequence tends to $l$, define $e_n=a_n-l$ and by substituting it in $a_{n+1}={1\over 4-a_n}$ conclude that $e_n \to 0$