Can somebody help me to prove, that sequence is decreasing $\textbf{without}$ using $x_{n+1}-x_n$ or $\frac{x_{n+1}}{x_n}$. For example if I have the sequence:
$a_1:=\sqrt{2}, a_{n+1}:=\sqrt{2+a+n}$ $\quad$ for $n \in \mathbb N$, then it's bounded $1<a_n<2$ and I can prove, that the sequnce is monotone increasing:
$a_n<2 \iff a_n^2<2\cdot a_n \iff a_n^2<a_n+a_n<2+a_n$ now we apply square root to get: $a_n<\sqrt{2+a_n}=a_{n+1}$.
So I want to prove that my sequence is monotone by this
equivalent transformations. But I'm stuck how to do it in my case. So I have
$x_1>1, x_{n+1}:=2-\frac{1}{x_n} \quad$ for $n \in \mathbb N$. It's bounded by $1\leq x_n \leq2$. This sequnce is decreasing, but how can we prove it? I started with:
$x_n \geq 1 \iff 1 \geq \frac{1}{x_n} \iff -1 \leq -\frac{1}{x_n}$ now add $2$ to both sides to get $1 \leq 2-\frac{1}{x_n}=x_{n+1}$ but the problem I "lost" my $x_n$ from the left side.
Can somebody say me, what I have to do?
Best Answer
I would suggest induction.
Notice that if $a<b$ then $-\dfrac{1}{a} < -\dfrac{1}{b}$.
Since $x_2 < x_1$, you know its true initially. Now assume $x_n < x_{n-1}$
Then $x_{n+1}=2-{1\over x_n} < 2-{1\over x_{n-1}}=x_n$.
Edit, without induction.
First notice that $y^2-2y+1=(y-1)^2\ge 0$. Specifically if $1 < y\le 2$ (your case) then this can be manipulated into
$$-\dfrac{y}{2y-1} < - \dfrac{1}{y}. \ \ \ \ \ \ \ \ \ (1)$$
So,
$$x_{n+1} = 2-\dfrac{1}{x_n} = 2-\dfrac{1}{2-\dfrac{1}{x_{n-1}}}=2-\dfrac{x_{n-1}}{2x_{n-1}-1}$$
$$ < 2-\dfrac{1}{x_{n-1}}=x_n \ \ \ \ \ \ \ \ \text{here we use (1)}.$$