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

Question

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.

Answer 1

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 $|a_n - \sqrt{2} | < \epsilon$.