1
$\begingroup$

Let $x_1> 1$ and let $x_{n+1} := 2 - \displaystyle\frac{1}{x{_n}}$ for $n \in \mathbb{ N}$. Show that $(x_n)$ is bounded and monotone. Find the limit. I am confused on how to show that the sequence is increasing or decreasing without having a specific value for $x_1$. I think that I should use induction, but how do I define my base case?

$\endgroup$
2
  • $\begingroup$ I would try something by induction. Use any value of $x_1 > 1$ just to gain some intuition if the sequence is increasing or decreasing. Once you have proved it converges, start with $x_{n+1} = 2 - \frac{1}{x_n}$ and pass the limit: $L = 2 - \frac{1}{L}$ and solve for $L$. $\endgroup$
    – Ivo Terek
    Jun 17, 2014 at 1:35
  • $\begingroup$ "How do I define my base case?" I think you don't. Leave everything in terms of $x_1$. The fact that $x_1 > 1 $ should be enough to conclude something. $\endgroup$
    – Ivo Terek
    Jun 17, 2014 at 1:37

4 Answers 4

2
$\begingroup$

Hint. To show that the sequence is always increasing, the base case would be that $x_2>x_1$, that is, $$2-\frac{1}{x_1}>x_1\ .$$ Similarly, to show it is always decreasing, you would need to start with $$2-\frac{1}{x_1}<x_1\ .$$ By doing a little algebra you should be able to work out which of these is correct.

$\endgroup$
1
$\begingroup$

Induction is a good plan. The statement to prove will be (if we assume we're going to prove it's increasing) "$\forall n\ge 1: x_n\le x_{n+1}$". The base case will be "$x_1\le x_2$". To prove this without knowing the value of $x_1$, there's really only one thing you can do: use the given recurrence to relate the value of $x_2$ to the value of $x_1$. That yields the statement $$ x_1 \le 2 - \frac1{x_1} $$ to be proved. So, solve this inequality, finding the values of $x_1$ for which it is true, and see if the problem's statement that $x_1>1$ lets you conclude that it's true.

$\endgroup$
1
$\begingroup$

Assume that $x_n>x_{n+1}$. Then invert, negate and sum $2$ to get $$2-\frac 1{x_n}>2-\frac 1{x_{n+1}}$$

which gives $x_{n+1}>x_{n+2}$. The boundedness is proved in a similar manner. What is a putative upper bound?

$\endgroup$
0
0
$\begingroup$

You can prove that $x_n \gt 1$ by using induction (this shows a lower bound).

To show monotonicity, you don't need induction.

$x_{n} - x_{n+1} = x_n + \frac{1}{x_n} - 2 \gt 0$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .