The sequence 1/n tends to zero

Example

Let $a_n = 1/n$ for each integer $n \ge 1$. Then $a_n \to 0$ as $n \to \infty$. Let $\varepsilon > 0$. By the Archimedean property of $\mathbb{R}$, there is a natural number $N$ with $N > 1/\varepsilon$. For each $n \ge N$ we get $|a_n - 0| = 1/n \le 1/N < \varepsilon$. By the epsilon-N definition, this shows $1/n \to 0$.

Remarks

The proof uses the Archimedean property, which says that the natural numbers are not bounded above in $\mathbb{R}$. With the algebra of limits, it gives many more limits, for example $1/n^2 \to 0$ and $(n + 1)/n \to 1$. Mathlib states it for $1/n$ over $\mathbb{R}$ as tendsto_one_div_atTop_nhds_zero_nat.

Explanations

L1 human

The numbers $1, 1/2, 1/3, 1/4, \dots$ get smaller and smaller, and they approach $0$. To get $1/n$ below $0.001$, it is enough to take $n > 1000$.