# Let $x_n$ be a sequence of integers such that $x_{k+1}\neq x_k$ holds for every $k\ge 1$. Show that $x_n$ is not convergent.

Question: Let $$x_n$$ be a sequence of integers such that $$x_{k+1}\neq x_k$$ holds for every $$k\ge 1$$. Show that $$x_n$$ is not convergent.

Solution: We will show that $$(x_n)_{n\ge 1}$$ is not a Cauchy sequence. To show that $$(x_n)_{n\ge 1}$$ is not a Cauchy sequence, it is enough to show that for some $$\epsilon >0$$ and for all $$N\in\mathbb{N}$$, there exists $$m,n\ge N$$ such that $$|x_m-x_n|\ge \epsilon$$.

Note that since $$x_{k+1}\neq x_k, \forall k\ge 1$$ and $$x_k\in\mathbb{Z}, \forall k\ge 1$$, implies that $$|x_{k+1}-x_k|\ge 1, \forall k\ge 1.$$

Thus let $$\epsilon=1$$ and fix any $$N\in\mathbb{N}$$. Next select $$m=N+1, n=N$$. Thus, $$|x_m-x_n|=|x_{N+1}-x_N|\ge 1=\epsilon.$$ Thus, $$(x_n)_{n\ge 1}$$ is not a Cauchy sequence, which implies that $$(x_n)_{n\ge 1}$$ is not convergent. Hence, we are done.

Is this proof correct and rigorous enough and is there any other way to solve the problem?

• If a sequence of integers is convergent It needs to be constant after some large enough n... – PAM1499 Sep 4 '20 at 16:49

A convergent sequence of integers will become constant after large enough n (for that just take $$\epsilon<1$$). Then It is obviously that It is not the case.