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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?

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

Edit: Your answer is correct.

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