Let $(a_n)_n$ be a convergent sequence of integers , what can we say about $(a_n)_n$?
(I don't understand what is meant by this question)
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1what is the question here? – Mohan Feb 23 '13 at 13:15
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3@Mohan: The OP is asking for help understanding what the question is asking. – Brian M. Scott Feb 23 '13 at 13:17
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1@Fayz: have you found an example of a convergent sequence of integers? – Emanuele Paolini Feb 23 '13 at 13:17
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HINT: Suppose that the sequence converges to some number $L$. Then for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $|a_n-L|<\epsilon$ whenever $n\ge m_\epsilon$. In particular, there is an $m_{1/4}\in\Bbb N$ such that $$|a_n-L|<\frac14$$ whenever $n\ge m_{1/4}$. Using the triangle inequality, we see that if $k,n\ge m_{1/4}$, then
$$|a_k-a_n|\le|a_k-L|+|L-a_n|<\frac14+\frac14=\frac12\;.$$
Now remember: each of these terms $a_n$ is an integer. What can you say about integers that are less than $\frac12$ apart?
Brian M. Scott
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@Nhoj_Gonk: I needed to ensure that $|a_n-L|<\frac12$ in order to ensure that $|a_k-a_n|<1$. Thus, I could have used any positive $\epsilon\le\frac12$. I just chose $\frac14$ because the resulting numbers are simple and make it obvious why $a_k$ must be equal to $a_n$. – Brian M. Scott Mar 06 '21 at 19:18
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@Nhoj_Gonk: I told you why I needed to make $|a_n-L|<\frac12$: to ensure that $|a_k-a_n|<1$. If you know that $a_k$ and $a_n$ are integers and that $|a_k-a_n|<1$, what can you conclude about $a_k$ and $a_n$? It’s something that you could not conclude if $|a_k-a_n|$ were $1$ or more. – Brian M. Scott Mar 07 '21 at 01:33
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@GoNK: That’s right: it turns out the such a sequence must be constant from some point on. – Brian M. Scott Mar 07 '21 at 02:34
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@GoNK: You’re welcome. Yes: in the notation in my answer we have $a_n=a_{m_{1/4}}$ for all $n\ge m_{1/4}$. – Brian M. Scott Mar 07 '21 at 02:43