Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Edit: Well, this is awkward, after asking the question (that did frustrate me for a few days) I was able to solve it myself without using (and even before looking at) the solutions provided. What am I supposed to do now?

Let $\{a_n\}_{n=1}^\infty$ a sequence converging to $L$. Prove that: c) If $\forall n\in\mathbb{N}$,$a_n\in\mathbb{Z}$ then $L\in\mathbb{Z}$

What I think I should do is: Assume for contradiction that $L\notin\mathbb{Z}$, Therefore exists $a\in\mathbb{Z}$ such that $a-1<L<a$. And since the sequence is converging I can choose an epsilon such that:


Therefore I need to find an epsilon such that:

$a-1\leq L-\epsilon<a_n<L+\epsilon\leq a$

But I really have no idea what to do here, I can only think of epsilons that are true for one side ($\epsilon=L-a+1$ or $\epsilon=a-L$)

share|cite|improve this question
In response to your edit: well done! The purpose of this site is it provide help with maths problems, and if you've found out how to do it on your own then that's even better than doing it with our hints (and much better than just taking one of our answers!) The usual protocol is to write an answer yourself, and accept that one. It seems slightly unusual I know, but it's an allowed and encouraged thing to do (see the faq). – Tom Oldfield Nov 25 '12 at 13:17


If $L\notin \mathbb Z$ then $\exists \varepsilon > 0 : |L-x|<\varepsilon \Rightarrow x \notin \mathbb Z$.

share|cite|improve this answer

Hint: prove that, under the given data, the sequence $\,\{a_n\}\,\subset \Bbb Z\,$ is eventually constant, meaning:

$$\{a_n\}\subset\Bbb Z\,\,,\,\,\lim_{n\to\infty}a_n=L\Longleftrightarrow\,\exists \,N\in\Bbb N\,\,s.t.\,\,n>N\Longrightarrow a_n=L$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.