Ceiling problem - Determining n0 for the convergence of 1/n

Question

Given that we have to prove that the sequence $ (a_n)_{n\geq1} $ for $ a_n = \frac 1 n $ converges, is this proof sufficient enough, especially in regards to determining the $n_0$?

Given that: $ \\~\\ \forall n \ge n_0 : |a_n - 0 | = \frac 1 n \leq \frac 1{n_0} < \epsilon \\ $

We can easily determine that: $ \\~\\ n_0 > \frac 1 \epsilon $

Can we now determine that $n_0$ is: $n_0 = \lceil \frac 1 \epsilon \rceil $?

The problem is that $\epsilon > 0$ which makes it possible to set $\epsilon = 1$ which would also mean that $n_0 = \lceil \frac 1 1 \rceil = 1$

That would mean $\frac 1 1 < 1 $, or did I understand something wrong?

Answer 1

For all intents and purposes, ceiling of $1/e$ is a fine answer, since for most limiting arguments, you care for small $\epsilon$; however, if it makes you happier, you can set $n_0 > \frac{1}{e} + 1$. The rest of the proof is fine.

Also, note that you don't have to find the smallest $n_0$ for the inequality to be true, you just need to find some $n_0$; so you could have equally said $500/\epsilon + 1000$

Answer 2

As you pointed out we cannot. I would put $n_0=\lfloor\frac{1}{\epsilon}\rfloor+1$.