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?

  • 1
    $\begingroup$ Refusing to "use $x_{n+1}-x_n$" for this sequence is whimsy. $\forall x>1\quad 2-\frac1x<x,$ period. $\endgroup$ Oct 22, 2022 at 18:57
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    $\begingroup$ What is the motivation behind avoiding these differences or ratios? Without this motivation, it's hard for us to decide on an approach that would qualify, since you may view it as "equivalent to" differences or ratios. $\endgroup$
    – Brian Tung
    Oct 22, 2022 at 19:08

1 Answer 1


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

$$-\dfrac{y}{2y-1} < - \dfrac{1}{y}. \ \ \ \ \ \ \ \ \ (1)$$


$$x_{n+1} = 2-\dfrac{1}{x_n} = 2-\dfrac{1}{2-\dfrac{1}{x_{n-1}}}=2-\dfrac{x_{n-1}}{2x_{n-1}-1}$$

$$ < 2-\dfrac{1}{x_{n-1}}=x_n \ \ \ \ \ \ \ \ \text{here we use (1)}.$$

  • $\begingroup$ Thank you for your answer. yes induction works good here, but can we do it without? $\endgroup$
    – mathguruu
    Oct 22, 2022 at 18:37
  • $\begingroup$ See the above edit. $\endgroup$
    – David P
    Oct 22, 2022 at 19:09
  • $\begingroup$ Note this is (in disguise and lengthy) the method in my comment above, using explicitely the "forbidden" $x_{n+1}-x_n$. But you are not blamable for this, David. $\endgroup$ Oct 22, 2022 at 21:07
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    $\begingroup$ @AnneBauval Reading your comment, I agree, and it can be shortened almost immediately. I tried to appeal to OP's request as best as possible. $\endgroup$
    – David P
    Oct 22, 2022 at 21:27

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