# An example of a sequence of rational numbers converging to an irrational number using the $\varepsilon-N$ definition.

I am trying to find a sequence of rational numbers that converges to an irrational number. But the difficulty I am facing is that I am required to show convergence of such a sequence using only the $$\varepsilon-N$$ definition of sequences. The definition is:

We say $$(a_n)\to a$$ if for every $$\varepsilon>0$$, there is an $$N\in\mathbb N$$ such that $$|a_n-a|<\varepsilon$$ for all $$n\ge N$$.

I know that $$(1+1/n)^n\to \mathrm e$$ is a very popular example. But I cannot see how to prvoe the convergence using the above definition. Any help/hints would be appreciated.

P.S.: I checked out quite a few MSE questions regarding this but none of those examples seem to provable by the above definition.

If you understand the general description of decimal approximations of real numbers then you can turn your understanding into a construction of examples. Let me do this for $$a = \sqrt{2}$$.
First define a sequence of integers: let $$m_n$$ be the greatest integer less than $$10^{n-1} \cdot \sqrt{2}$$. While the actual values aren't needed for the proof, one knows of course that $$m_1 = 1, \qquad m_2 = 14, \qquad m_3 = 141, \qquad m_4 = 1414 \ldots$$ Notice that $$m_n < 10^{n-1} \sqrt{2} < m_n+1$$ Next define a sequence of rational numbers: $$a_n = \frac{m_n}{10^{n-1}}$$. Notice that $$a_n < \sqrt{2} < a_n + \frac{1}{10^{n-1}}$$ It follows that $$| a_n - \sqrt{2} | < \frac{1}{10^{n-1}}$$ The $$\epsilon$$-$$N$$ argument is easily completed: for each $$\epsilon > 0$$ choose an integer $$N > \log_{10}(1/\epsilon)+1$$, and it follows that if $$n \ge N$$ then $$|a_n - \sqrt{2} | < \epsilon$$.