Given a recurrence relation
$$
x_{n+1} = \frac{1}{2x_n – 1}\\
x_1 = {9\over 10}\\
n\in\Bbb N
$$
Figure out whether the sequence converges and find $\lim x_n$ in case it exists.
I've started with finding a fixed point of the recurrence, namely:
$$
L = \frac{1}{2L -1} \iff 2L^2-L -1 = 0 \iff (L-1)\left(L+{1\over 2}\right) = 0
$$
The only possible finite limits are $\left\{1, -{1\over 2}\right\}$.
Then I proceeded with calculating a few first terms:
$$
x_n = \left\{{9\over 10}, {5\over 4}, {2\over 3}, 3, {1\over 5}, -{5\over 3}, -{3\over 13}, \cdots \right\}
$$
Using induction suppose that:
$$
\forall n\ge6:x_n \le 0
$$
We have:
$$
x_n < 0 \implies 2x_n < 0\implies 2x_n – 1 < 0 \implies \underbrace{{1\over 2x_n -1}}_{x_{n+1}} < 0
$$
So it follows that indeed all terms after $6$-th are less than $0$. And we are left with only one possible value for the limit, namely $L = -{1\over 2}$.
At this point I got into troubles. The problem is that according to the graph
the sequence is bouncing around its limit point.
A reasonable step would be to prove that $x_{2n}$ is monotonically increasing while $x_{2n-1}$ is monotonically decreasing and both are bounded by $-{1\over 2}$ above and below accordingly, but I failed to do that. Is there a simple way to prove monotonicity? If not what steps should I take to finish the study of the sequence?
I was thinking about induction, but then I have to "jump" over odd and even indices for $x_{2n}$ and $x_{2n-1}$, not sure whether that's applicable. The equations for odd and even terms are as follows:
$$
x_{2n} = \frac{2x_{2n-2}-1}{3-2x_{2n-2}}\\
x_{2n-1} = \frac{2x_{2n-3}-1}{3-2x_{2n-3}}
$$
What I want to prove eventually is:
$$
x_{2n-2} \le x_{2n} \le -{1\over 2} \\
x_{2n-1} \ge x_{2n-3} \ge -{1\over 2}
$$
Please note that this question is from limits section. Even though it describes some kind of a dynamic system I would like to keep away from that part of Math since I'm not familiar with it, and use concepts understandable by pre-calc/calc 1 student.
Best Answer
If $f(x) = 1/(2 x - 1)$ we have $$ \frac{f(x) + 1/2}{x + 1/2} = \frac{1}{2x-1} $$ Now $$ \left| \frac{1}{2x-1} \right| < 1 \ \text{if}\ x < 0$$ Thus as soon as you get negative values of $x_n$, $|x_n + 1/2|$ decreases in each iteration. Moreover, as you mentioned, the $x_n$ stay negative. Therefore $\lim_{n \to \infty} |x_n + 1/2|$ must be $0$, i.e. $\lim_{n \to \infty} x_n = -1/2$.