Real numbers/1 over 3/Sequence of decimal expansions/Example

We consider the sequence

${}x_{n}=0.33\ldots 33\,,$

with exactly ${}n$ digits after the point. We claim that this sequence converges to ${}1/3$. For this, we have to determine ${}\vert {0,33\ldots 33-{\frac {1}{3}}}\vert$, and before we can do this, we have to recall the meaning of a decimal expansion. We have

${}x_{n}=0.33\ldots 33={\frac {33\ldots 33}{10^{n}}}={\frac {\sum _{j=0}^{n-1}3\cdot 10^{j}}{10^{n}}}\,,$

and therefore

${}{\begin{aligned}\vert {0,33\ldots 33-{\frac {1}{3}}}\vert &=\vert {{\frac {\sum _{j=0}^{n-1}3\cdot 10^{j}}{10^{n}}}-{\frac {1}{3}}}\vert \\&=\vert {\frac {3\cdot {\left(\sum _{j=0}^{n-1}3\cdot 10^{j}\right)}-10^{n}}{3\cdot 10^{n}}}\vert \\&=\vert {\frac {{\left(\sum _{j=0}^{n-1}9\cdot 10^{j}\right)}-10^{n}}{3\cdot 10^{n}}}\vert \\&=\vert {\frac {-1}{3\cdot 10^{n}}}\vert \\&={\frac {1}{3\cdot 10^{n}}}.\end{aligned}}$

If now a positive ${}\epsilon$ is given, then for ${}n$ sufficiently large, this last term is ${}\leq \epsilon$.