I'm trying to prove from the that the following sequence converges to $\sqrt{2}$, even though it seems obvious.
We define $x_n = \frac{m_n}{10^n}$ where $m_n$ is defined as the largest natural number such that $\left(\frac{m_n}{10^n}\right)^2 < 2$. After plugging in some values, I find
$$
(x_n) = \{1.4, 1.41, 1.414, 1.4142, \ldots \}
$$
and the idea is that $x_n$ agrees with $\sqrt{2}$ up to the $n$th decimal position. The problem is that I don't "know" that by the construction. If I did, I could write out the decimal expansion of $\sqrt{2}$ and find $N$ such that $\frac{1}{10^N} < \epsilon$ by the Archimedean property. Then taking $n \geq N + 1$, we have $|x_n – \sqrt{2}| < \frac{1}{10^N} < \epsilon$. But I don't believe I know at the start of the problem that this actually is the decimal expansion of $\sqrt{2}$, so this proof assumes the conclusion. I can prove that the sequence is Cauchy and hence convergent, but I can't prove that the limit is actually $\sqrt{2}$.
Best Answer
If $m_n$ is the largest integer such that $\left(\frac{m_n}{10^n}\right)^2 < 2$ then $$ \left(\frac{m_n}{10^n}\right)^2 < 2 \le \left(\frac{m_n+1}{10^n}\right)^2 $$ which implies $$ x_n < \sqrt 2 \le x_n + 10^{-n} $$ or $$ \sqrt 2 - 10^{-n} \le x_n < \sqrt 2 \, . $$ $x_n \to \sqrt 2$ now follows, e.g. by the squeeze theorem.
Actually all inequalities are strict since $x_n$ is rational and $\sqrt 2$ is irrational, but that is not needed in order to prove the convergence.