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

• @Saad there's a dot in his/her name – mathworker21 Apr 24 at 4:25

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.