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let $x_{1}=0,x_{2}=1$,for $n\ge 2$,we have $$x_{n+1}=\left(2-\dfrac{1}{n}\right)x_{n}-x_{n-1}$$ show that: there exist postive integer $N$, such for all $n\ge N$,we have $$x_{n}x_{n+1}\le 0$$

My try: if $x_{n}x_{n+1}>0,\forall n\ge N$,since $$x_{n}x_{n+1}=\left(2-\dfrac{1}{n}\right)x^2_{n}-x_{n-1}x_{n}$$so $$x_{n}x_{n+1}+x_{n-1}x_{n}=\left(2-\dfrac{1}{n}\right)x^2_{n}$$

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  • $\begingroup$ @ApassJack's answer has disproved your proposition. $\endgroup$ – Saad Apr 24 at 2:46
  • $\begingroup$ @Saad there's a dot in his/her name $\endgroup$ – mathworker21 Apr 24 at 4:25
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On the contrary, there does not exist a positive integer $N$ such that for all $n\ge N$, we have $$x_{n}x_{n+1}\le 0.$$

Proof: Suppose there is such $N$. Let $n= N$. If $x_n = 0$, then $x_{n+1}\not=0$; otherwise, with the condition $x_n=x_{n+1}=0$, we will have $x_{n-1}=0$, $x_{n-2}=0$, $\cdots$, and, in the end, $x_1=0$, which is not true. In that case, we will set $n=N+1$. We will always have $x_n\not=0$.

There are two cases. Note that $2-\frac{1}{m+1}\gt1$ for all positive integer $m$.

  • $x_n>0$. Then $x_{n+1}\le 0$. We have, $$x_{n+2}=\left(2-\frac{1}{n+1}\right)x_{n+1}-x_{n}\le x_{n+1}-x_{n}\lt 0.$$ $$x_{n+3}=\left(2-\frac{1}{n+2}\right)x_{n+2}-x_{n+1}\le x_{n+2}-x_{n+1}\le (x_{n+1}-x_{n})-x_{n+1}=-x_n<0.$$ So, $x_{n+2}x_{n+3}>0$.

  • $x_n<0$. Then $x_{n+1}\ge 0$. We have, $$x_{n+2}=\left(2-\frac{1}{n+1}\right)x_{n+1}-x_{n}\ge x_{n+1}-x_{n}\gt 0.$$ $$x_{n+3}=\left(2-\frac{1}{n+2}\right)x_{n+2}-x_{n+1}\ge x_{n+2}-x_{n+1}\ge (x_{n+1}-x_{n})-x_{n+1}=-x_n\gt0.$$ So, $x_{n+2}x_{n+3}>0$. (Yes, this case is symmetric to the case of $x_n>0$.)

So, in all cases, we have found $x_{n+2}x_{(n+2)+1}>0$, which contradicts our assumption.


It is, in fact, immediate to show that the sign of the elements can not change (strictly) twice in a roll.

$$x_{n+1}+x_{n-1}=\left(2-\dfrac{1}{n}\right)x_{n}$$

Assume none of $x_{n+1}$, $x_{n-1}$, $x_{n}$ is zero. Then one of $x_{n+1}$ and $x_{n-1}$ must have the same sign as $x_{n}$; otherwise, the LHS will have the opposite sign with the RHS.

With some fine argument, we can prove that $x_n\not=0$ for all $n\not=1$. Then we can see that the sequence never flips its sign consecutively.


An interesting question might be how to show that sequence will change its sign infinitely many times.

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