We consider the sequence
-
![{\displaystyle {}x_{n}=0.33\ldots 33\,,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/473b4b1dc437d9563ffbeec857901f6d57567421)
with exactly
digits after the point. We claim that this sequence converges to
. For this, we have to determine
, and before we can do this, we have to recall the meaning of a decimal expansion. We have
-
![{\displaystyle {}x_{n}=0.33\ldots 33={\frac {33\ldots 33}{10^{n}}}={\frac {\sum _{j=0}^{n-1}3\cdot 10^{j}}{10^{n}}}\,,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7efdcd8f0d4d45fa711f764db93f96b42bb66924)
and therefore
![{\displaystyle {}{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/671c6847f4ddb95d4c51fcf6e7f8caf9cf25b61a)
If now a positive
is given, then for
sufficiently large, this last term is
.