Answer
$$\frac{1}{9}$$
Work Step by Step
$$\eqalign{
& \sum\limits_{k = 1}^\infty {{{10}^{ - k}}} \cr
& {\text{Rewriting}} \cr
& \sum\limits_{k = 1}^\infty {{{10}^{ - k}}} = \sum\limits_{k = 1}^\infty {{{\left( {\frac{1}{{10}}} \right)}^k}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sum\limits_{k = 1}^\infty {\left( {\frac{1}{{10}}} \right){{\left( {\frac{1}{{10}}} \right)}^{k - 1}}} \cr
& {\text{,Then }}{a_n} = \underbrace {\frac{1}{{10}}{{\left( {\frac{1}{{10}}} \right)}^{k - 1}}}_{{a_1}{r^{n - 1}}} \cr
& {\text{The sum is }}{S_\infty } = \frac{{{a_1}}}{{1 - r}} \cr
& \sum\limits_{k = 1}^\infty {{{10}^{ - k}}} = \frac{{1/10}}{{1 - 1/10}} \cr
& {\text{Simplifying}} \cr
& \sum\limits_{k = 1}^\infty {{{10}^{ - k}}} = \frac{1}{9} \cr} $$
