Infinite sum of random variables 
Let $\xi_i$ be independent discrete random variables such that $P[\xi_i =    k] = {1 \over 10}$ for all $i \in \mathbb{N}$ and $k = 0,\dots,9$. Define
$$
X = \sum_{i = 1}^\infty \xi_i{1 \over 10^i}.
$$
What's the distribution of $X$?

The sum converges as every summand is less than ${10 \over 10^i} = {1 \over 10^{i-1}}$ and $\sum_{i=1}^\infty {1 \over 10^i}$ converges.
I let $X_i = {\xi_i\over 10^i}$. Then I have $X = \sum_{i = 1}^\infty X_i$. From this formula it's easy to see that every $X_i \in \{0, {1\over 10^i}, {2\over 10^i}, \dots, {9\over 10^i}\}$, ie. the $i$-th random variable contributes only to the $i$-th decimal place of the sum $X$ and does not affect other decimal places.
This immediately gives $X \in [0, 1)$. Also I think that $P[X=x] = 0$ for any $x$ in $[0, 1)$, suggesting "continuous behaviour" of $X$.
Could anyone provide hint which way to go from here? Maybe the Central limit theorem or Law of large numbers could give something.
Thanks.

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For a given $\ds{N \in \mathbb{N}_{\geq 1}}$, the $\ds{X}$-distribution is given by
  $\ds{\pars{~\mbox{note that, in this case,}\ 0 \leq X \leq 10 - {1 \over 10^{N}}~}}$:

\begin{align}
&\left.\sum_{\xi_{1} = 0}^{9}{1 \over 10}\cdots\sum_{\xi_{N} = 0}^{9}{1 \over 10}\right\vert_{\ X\ =\ \sum_{i = 1}^{N}\xi_{i}/10^{i}} =
{1 \over 10^{N}}\sum_{\xi_{1} = 1}^{9}\cdots\sum_{\xi_{N} = 1}^{9}
\bracks{z^{X}}z^{\sum_{i = 1}^{N}\xi_{i}/10^{i}}
\\[5mm] = &\
{1 \over 10^{N}}\sum_{\xi_{1} = 1}^{9}\cdots\sum_{\xi_{N} = 1}^{9}
\bracks{z^{X}}\pars{z^{\xi_{1}/10}z^{\xi_{2}/100}\cdots z^{\xi_{N}/10^{N}}}
\\[5mm] = &\
{1 \over 10^{N}}\bracks{z^{X}}\pars{%
\sum_{\xi_{1} = 0}^{9}z^{\xi_{1}/10}\sum_{\xi_{2} = 0}^{9}z^{\xi_{2}/100}
\ldots\sum_{\xi_{N} = 0}^{9}z^{\xi_{N}/10^{N}}}
\\[5mm] = &\
{1 \over 10^{N}}\bracks{z^{X}}\pars{{z - 1 \over z^{1/10} - 1}
{z^{1/10} - 1 \over z^{1/100} - 1}\ldots
{z^{1/10^{N - 1}} - 1 \over z^{1/10^{N}} - 1}} =
{1 \over 10^{N}}\bracks{z^{X}}\pars{1 - z \over  1 - z^{1/10^{N}}}
\\[5mm] = &\
{1 \over 10^{N}}\bracks{z^{X}}\pars{1 \over  1 - z^{1/10^{N}}} -
{1 \over 10^{N}}\bracks{z^{X - 1}}\pars{1 \over  1 - z^{1/10^{N}}}
\\[5mm] = &\
{1 \over 10^{N}}\bracks{z^{X}}\sum_{k = 0}^{\infty}z^{k/10^{N}} -
{1 \over 10^{N}}\bracks{z^{X - 1}}\sum_{k = 0}^{\infty}z^{k/10^{N}} =
\mrm{f}_{N}\pars{X} - \,\mrm{f}_{N}\pars{X - 1}
\\ &\
\bbx{\ds{\mbox{where}\ 0 \leq X \leq 10 - {1 \over 10^{N}}}}
\end{align}

$$
\mbox{and}\quad
\mrm{f}_{N}\pars{x} \equiv
\left\{\begin{array}{lcl}
\ds{1 \over 10^{N}} & \mbox{if} & \ds{10^{N}x\ \in\ \mathbb{N}_{\geq 0}}
\\[2mm]
\ds{0} && \mbox{otherwise}
\end{array}\right. 
$$
