Sequences and the ε–N Limit

A sequence $(a_n)$ is an ordered, infinite list of real numbers: $a_1, a_2, a_3, \dots$. We say $(a_n)$ converges to a limit $L$, written $\lim_{n\to\infty} a_n = L$, when the terms get and stay arbitrarily close to $L$. Made precise, this is the $\varepsilon$–$N$ definition:

Read it as a challenge–response game. Someone hands you a tolerance $\varepsilon$ (however tiny). You must find a cut-off $N$ so that every term past $a_N$ sits within $\varepsilon$ of $L$. If you can always answer, $L$ is the limit.

Worked example. Prove that $a_n = \dfrac{1}{n} \to 0$.

Given any $\varepsilon > 0$, choose a natural number $N > \dfrac{1}{\varepsilon}$. Then for every $n > N$,

Because we could meet any $\varepsilon$, the limit is $0$. Notice we never needed $a_n$ to actually equal $0$ — closeness that we can force as tight as we like is enough.