By inspection, it's clear that the sequence must converge to because
when is arbitrarily large.
Now, for the limit as to be equal to is to say that for any , there exists some such that whenever , it follows that
From this inequality, we get
As we're considering , we can omit the first inequality.
We can then see that choosing will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that .