The question states:

There are $n$ positive numbers $x_1,x_2,\ldots, x_n$, where $n \geq 3$, satisfy $$ x_1=1+\frac{1}{x_2},\ x_2 = 1+\frac{1}{x_3},\ \cdots, \ x_{n-1}=1+\frac{1}{x_n} $$ and also, $$ x_n=1+\frac{1}{x_1} $$ Show that

  1. $x_1\cdot x_2\cdot x_3 \cdots x_n > 1$
  2. $x_1 - x_2 = -\frac{x_2-x_3}{x_2x_3}$
  3. $x_1 = x_2 = \cdots = x_n$

Hence find the value of $x_1$

I could do $(1)$ and $(2)$ quite easily. Here's how:

  1. Plugging in $$ x_1=1+\frac{1}{x_2},\ x_2 = 1+\frac{1}{x_3},\ \cdots, \ x_{n-1}=1+\frac{1}{x_n} $$ and $$ x_n=1+\frac{1}{x_1} $$

in $x_1\cdot x_2\cdot x_3 \cdots x_n$, we get $$ \left(1+\frac{1}{x_1}\right)\cdot \left(1+\frac{1}{x_2}\right)\cdot \left(1+\frac{1}{x_3}\right)\cdots \left(1+\frac{1}{x_n}\right)\\ =1+k>1 $$ where $k$ is a positive number.

  1. Just by substituting the expressions for $x_1$ and $x_2$, we get $$ x_1-x_2=1+\frac{1}{x_2}-1-\frac{1}{x_3}=\frac{1}{x_2}-\frac{1}{x_3}\\ =-\frac{x_2-x_3}{x_2x_3} $$

Now I've tried for a day but I couldn't find a way to prove that they're all equal. A nudge in the proper direction would be appreciated.


Following copper.hat's advice, I decided to analyse the function $f(x)=1+\frac{1}{x}$. Here's what I did:

$$ f(x)=1+\frac{1}{x} \\ (f \circ f)(x)=f^2(x)=1+\frac{1}{1+\frac{1}{x}}\\ \vdots\\ \lim_{n \to \infty}\underbrace{f \circ f \cdots f}_{n \text{ times}}=\lim_{n \to \infty}f^n(x)=1+\frac{1}{1+\frac{1}{\ddots}}=\phi $$

Therefore, intuitively it's easy to see that all the terms are $\phi$. But how do I prove this rigorously? Or how should I write it to prove the third claim? Basically this post is about seeking guidance in mastering Mathematical lingo now.

  • 2
    $\begingroup$ Let $f(x) = 1+ {1 \over x}$ and examine how $f(f(...f(x)...))$ behaves and look for fixed points. For $x \ge 1$, $f$ is bounded and monotonic. $\endgroup$ – copper.hat Feb 3 '17 at 6:25
  • $\begingroup$ @copper.hat It seems to be behaving sort of like $f^n(x)=\frac{F_n x+F_{n-1}}{F_{n-1}x+F_{n-2}}$, where $F_n$ is the $n$th Fibonacci number. $\endgroup$ – Hungry Blue Dev Feb 3 '17 at 6:43
  • $\begingroup$ @copper.hat Am I correct? $\endgroup$ – Hungry Blue Dev Feb 3 '17 at 6:46
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    $\begingroup$ @Astrobleme Not too surprising, since if $x_1=x_2$ then it's the root of $x^2-x-1=0$ i.e. the golden ratio. $\endgroup$ – dxiv Feb 3 '17 at 6:49
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    $\begingroup$ For each $i$ we have $|x_i-x_{i+1}|=|(x_{i+1}-x_{i+2})/(x_ix_{i+1})|\leq|x_{i+1}-x_{i+2}|$, because $x_j>1$ for each $j$. $\endgroup$ – Matemáticos Chibchas Feb 3 '17 at 7:44

Using $(2)$ repeatedly you get $$ \lvert x_1 - x_2 \rvert = \frac{\lvert x_2-x_3\rvert }{x_2x_3} = \frac{\lvert x_3-x_4\rvert }{x_2x_3^2x_4} = \ldots = \frac{\lvert x_{n}-x_1 \rvert }{x_2x_3^2x_4^2\cdots x_{n-1}^2x_n^2x_1} \\ = \frac{\lvert x_1-x_2 \rvert }{x_2x_3^2x_4^2\cdots x_{n-1}^2x_n^2x_1^2 x_2} = \frac{\lvert x_1-x_2 \rvert }{(x_1\cdots x_n)^2} $$ In $(1)$ you have shown that the denominator is strictly greater than one, therefore $x_1 - x_2 = 0$ must hold, and consequently, all $x_i$ are equal.


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