For all $x\in\mathbb{R}$, let $\lbrack x\rbrack$ denote the greatest integer $b\in\mathbb{Z}$ such that $b\leq x$. I would like to solve the following problem.
Define a sequence $a_n=\lbrack 5/n\rbrack$ for all $n\in\mathbb{N}$. Using the definition of convergence, prove that $\lim a_n=0$.
My Attempt: To show that $\lim a_n=0$ is to show that $|\lbrack 5/n\rbrack|<\epsilon$ for all $\epsilon >0$. Observe that for each $n\in\mathbb{N}$, $|\lbrack 5/n\rbrack|\leq 5/n <6/n$. Then, we can choose $n_0\in\mathbb{N}$ such that $n_0>6/\epsilon$ by Archimedean property. For any $n\geq n_0$, we have $$n>6/\epsilon\implies 6/n<\epsilon\implies |\lbrack 5/n\rbrack|<\epsilon$$ which completes the proof.
It seems right, but I'm not sure if each of these steps are "allowed". Would appreciate if someone could confirm/correct my proof.
You're making it more complicated than it needs to be (as is Fred in the other answer). What are the elements of the sequence for $n>5$? Now use the definition of convergence on that instead.