there exist postive integer $N$, such for all $n\ge N$,we have $x_{n}x_{n+1}\le 0$ 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
$$x_{n}x_{n+1}+x_{n-1}x_{n}=\left(2-\dfrac{1}{n}\right)x^2_{n}$$
 A: 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 $m$. 


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*$x_n>0$. 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.
