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

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

## 5 Answers

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 $$x \in (2-\sqrt3,2+\sqrt3)$$.

• Why the downvotes? Because this post explains a simple, illuminating, rigorous, graphical approach which avoids basically any formula and is definitely worth knowing?
– Did
Commented Dec 2, 2018 at 10:37
• You mean, your answer was downvoted because you (correctly) pointed out a serious flaw in another post? This would be outrageous.
– Did
Commented Dec 2, 2018 at 10:42
• It got edited before people saw the flaw I pointed. That would be the only reason. Commented Dec 2, 2018 at 10:44

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.

• 1. The formula for $g'(x)$ is wrong. 2. Computing $g'(x)$ is not necessary at all to solve the question. (But two upvotes and an accepted answer in 1 hour...)
– Did
Commented Dec 2, 2018 at 9:21
• Since my first comment, a silent correction of $g'(x)$ occurred. Point 2. still stands, fully.
– Did
Commented Dec 2, 2018 at 10:59
• "This comment is rude or condescending. Learn more in our Code of Conduct."
– Did
Commented Dec 2, 2018 at 21:22

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 $$p(0)=1, p(2)=-3$$

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 $$y$$.

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 by using induction.

• You do not need to invoke $2+\sqrt{3}$. Use $x_1=3$ as the upper bound and accept $\le$ for the comparison with this bound. Commented Dec 2, 2018 at 10:49
• Yes, i realized it later. Thanks :). Commented Dec 2, 2018 at 15:11

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 $$e_n \to 0$$