# Prove the sequence $x_{n+1}=\frac{1}{4-x_n}, x_1=3$ converges.

I try to use induction to show that the sequence is monotonically decreasing. My induction hypothesis is $$P(n):x_{n+1} We have first few terms $$x_1=3, x_2=1, x_3=1/3,\cdots$$. It is clear that $$P(1)$$ and $$P(2)$$ hold. Now I try to prove $$P(n+1)$$ assuming $$P(n)$$ is true. To do this I must show that $$x_{n+2}-x_{n+1}<0$$.

$$x_{n+2}-x_{n+1}=\frac{1}{4-x_{n+1}}-\frac{1}{4-x_n}=\frac{x_{n+1}-x_n}{(4-x_{n+1})(4-x_n)}$$ I get stuck here I know that numerator is less than $$0$$ but how do I show that denominator is positive so that the entire thing is less than $$0$$.

I am trying to show that sequence is monotonically decreasing and also bounded below to show that it converges.

Hints first

• First hint: the function $x\to\dfrac1{4-x}$ is monotone increasing (as long as $x<4$, naturally). – user436658 Feb 20 '18 at 9:00
• @ProfessorVector I don't want to use notions of functions as they haven't been introduced yet in the book I am following. Is there a way to do this using only properties of sequences. – Sonal_sqrt Feb 20 '18 at 9:07
• There's only one way to do mathematics: using mathematical notions. If you don't want that, you'd better ask your questions elsewhere, sorry. – user436658 Feb 20 '18 at 9:09
• @ProfessorVector I can totally see that your method would work, but there are more than one way to solve any problem. – Sonal_sqrt Feb 20 '18 at 9:16
• Then, it's all the more sad that you didn't find any. – user436658 Feb 20 '18 at 9:17

Hint: use as your inductive hypothesis the following statement: $x_{n+1} < x_n < 4$.

How to come up with this hypothesis? I first asked myself whether you were asking for something that's even true: is the denominator positive? The answer is obviously yes by working out the first few values. Why is the denominator positive? Because it's the product of positive things. Can we show that we're always taking the product of positive things? In order to do that, we'd need…

• this bit proves the sequence monotonic, I also need to prove that there is some lower bound so that I can conclude that it converges. Right? – Sonal_sqrt Feb 20 '18 at 9:13
• @PiyushDivyanakar can the terms ever be negative? that should give you a lower bound :) – Zubin Mukerjee Feb 20 '18 at 9:14
• Ah, I assumed you only wanted a hint for the "prove that the sequence is monotone" part. But see ZubinMukerjee's comment. – Patrick Stevens Feb 20 '18 at 9:14
• Thanks a lot @ZubinMukerjee and Patrick, I see that all lower bound is already implied by the induction hypothesis. – Sonal_sqrt Feb 20 '18 at 9:22
• (If you want to be totally precise, you could either change the inductive hypothesis to $0 < x_{n+1} < x_n < 4$, or you could do another induction at the end to show that $x_n > 0$.) – Patrick Stevens Feb 20 '18 at 9:24

We can realize easily:

$x_{n+1}=\frac{1}{4-x_n}$ = $\frac{1}{4}\sum_{k=0}^{\infty}\big(\frac{x_n}{4}\big)^k$

and

$x_n=\frac{1}{4}\sum_{k=0}^{\infty}\big(\frac{x_{n-1}}{4}\big)^k$

Using induction method we know that $x_2\lt x_1$ and if $x_{n}\lt x_{n-1}$ then $x_{n+1}\lt x_{n}$ can be proved.

Minorize $x_n=\frac{1}{4}\sum_{k=0}^{\infty}\big(\frac{x_{n-1}}{4}\big)^k$ with the substitution of $x_{n-1}$ by $x_{n}$ then

$x_n\gt\frac{1}{4}\sum_{k=0}^{\infty}\big(\frac{x_{n}}{4}\big)^k=x_{n+1}$

• For lower bound: $x_n=4-\frac{1}{x_{n+1}}$ and if $x_n\gt x_{n+1}$ then $x_n\lt4-\frac{1}{x_{n}}$, after rearranging we get that $(x_n-2)^2\lt 3$ Solve the unequality we have that: $2-\sqrt3 \lt x_n\lt2+\sqrt3$ – JV.Stalker Feb 20 '18 at 12:40