I am having a lot of trouble with the concept of proving the limit of a sequence using epsilon n. As an example, I am working on trying to prove that the limit of the sequnce $\displaystyle \frac{n+1}{2n}$ is $\displaystyle \frac{1}{2}$. I understand that the first step is to set $\displaystyle \frac{n+1}{2n} - \frac{1}{2} < \epsilon$. I then solve for n and get $\displaystyle n>\frac{1}{2\epsilon}$. So I think I understand this tells me that for any $\displaystyle \epsilon$, picking an $\displaystyle n>\frac{1}{2\epsilon}$ will give a value less than $\displaystyle \epsilon$. I'm just not sure where to go next with the proof. I've arbitraily tried different values than $\displaystyle \frac{1}{2}$ for the limit and every time I can solve $\displaystyle \frac{n+1}{2n} -$ arbitrary number for $\displaystyle n > $ something with $\displaystyle \epsilon$ in it. So I must be doing something wrong as it seems using my method I can prove the limit to be any arbitrary number. In searching online and other forums it seems there is maybe a last step where I take my $\displaystyle n>\frac{1}{2\epsilon}$ and somehow work in the "other direction" to prove that indeed any $\displaystyle n>\frac{1}{2\epsilon}$ works, but I've been unable to follow. Thanks for any help.