I was trying to look at the problem:
Given that:
There are n positive number x_1, x_2, …., x_n, where n>= 3 satisfy:
$x_1 =1+\frac{1}{x_2}， x_2 = 1+\frac{1}{x^3} ,…, x_{n-1} = 1+\frac{1}{x_n}$, and also,$ x_n = 1 + \frac{1}{x_1}$.
Prove $x_1 = x_2 = x_3 = … = x^n$
This problem previous part has shown that:
all x values bigger than 1,and $x_1 – x_2 = -\frac{x_2 – x_3}{x_2 x_3}$
This question is taken from a book, and the book has proven $x_1 = x_2$ to me, but I don't know how to continue.
Thank you so much for you guy's replies.
Best Answer
For all $1\leq k<n-1$ observe that $$ x_{k+1}-x_k=\frac{1}{x_{k+2}}-\frac{1}{x_{k+1}}=\frac{x_{k+1}-x_{k+2}}{x_{k+1}x_{k+2}}, $$ and therefore $$ |x_1-x_2|=\frac{|x_2-x_3|}{|x_2x_3|}=\frac{|x_3-x_4|}{|x_2x_3^2x_4|}=\cdots=\frac{|x_{n-1}-x_n|}{|x_2x_3^2x_4^2\cdots x_{n-1}^2x_n|}. $$ But from the last equation $x_n=1+\tfrac{1}{x_1}$ you also have that $$ x_n-x_{n-1}=\frac{1}{x_1}-\frac{1}{x_n}=\frac{x_n-x_1}{x_nx_1} $$ and $$ x_1-x_n=\frac{1}{x_2}-\frac{1}{x_1}=\frac{x_1-x_2}{x_1x_2}, $$ so combining it all together gives you $$ |x_1-x_2|=\frac{|x_1-x_2|}{|x_1^2x_2^2x_3^2\cdots x_{n-1}^2x_n^2|}. $$ The only way this is true is if $|x_1-x_2|=0$, since the denominator is always greater than $1$. And from the proof, we see that $|x_2-x_3|=\cdots=|x_{n-1}-x_n|=0$ as well, so all the $x_k$ are equal.