Epsilon-N definition of the limit of a sequence

Statement

Let $(a_n)$ be a real sequence and let $L$ be a real number. Then $(a_n)$ converges to $L$, in the sense of the limit of a sequence, if and only if the following holds. For each real number $\varepsilon > 0$ there is a natural number $N$ such that $|a_n - L| < \varepsilon$ for all $n \ge N$.

Remarks

The proof uses two facts. Each interval $(L - \varepsilon, L + \varepsilon)$ is open and contains $L$. Each open interval that contains $L$ contains such an interval. In a metric space $(X, d)$ the same statement holds with $d(a_n, L)$ in place of $|a_n - L|$. The number $N$ depends on $\varepsilon$. A smaller $\varepsilon$ in general needs a larger $N$.

Explanations

L1 human

The number $\varepsilon$ is the allowed error, and $N$ is the point after which the error is always smaller than $\varepsilon$. For $a_n = 1/(n+1)$ and $L = 0$, the choice $N > 1/\varepsilon$ works.