Problem source

\begin{questionparts}
\item By using the formula for the sum of a geometric series, or
otherwise, express the number $0.38383838\ldots$ as a fraction in
its lowest terms. 
\item Let $x$ be a real number which has a recurring decimal expansion
\[
x=0\cdot a_{1}a_{2}a_{2}\cdots,
\]
so that there exists positive integers $N$ and $k$ such that $a_{n+k}=a_{n}$
for all $n>N.$ Show that 
\[
x=\frac{b}{10^{N}}+\frac{c}{10^{N}(10^{k}-1)}\,,
\]
where $b$ and $c$ are integers to be found. Deduce that $x$ is
rational. 
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$ \begin{align*}
&& 0.383838\ldots &= \frac{3}{10} + \frac{8}{100} + \frac{3}{1000} + \cdots \\
&&&= \frac{38}{100} + \frac{38}{10000} + \cdots \\
&&&= \frac{38}{100} \left (1 + \frac{1}{100} + \frac{1}{100^2} + \cdots  \right) \\
&&&= \frac{\frac{38}{100}}{1 - \frac{1}{100}} \\
&&&= \frac{38}{99}
\end{align*}

\item Let $x = 0\cdot a_{1}a_{2}a_{2}\cdots$ such that there exists $N$, $k$ such that $a_{n+k} = a_n$ for $n > N$.

First notice that

\begin{align*}
x &= \frac{a_1a_2\cdots a_N}{10\ldots000} + \frac{a_{N+1}}{10^{N+1}} + \cdots \\
&= \frac{b}{10^N} + \frac{a_{N+1}a_{N+2}\cdots a_{N+k}}{10^{N+k}} + \frac{a_{N+1}a_{N+2}\cdots a_{N+2k}}{10^{N+k}} + \cdots \\
&= \frac{b}{10^N} + \frac{c}{10^{N+k}} \left (1 + \frac{1}{10^k} + \cdots \right) \\
&=  \frac{b}{10^N} + \frac{c}{10^{N+k}} \frac{1}{1- \frac{1}{10^k}} \\
&=  \frac{b}{10^N} + \frac{c}{10^{N+k}} \frac{10^k}{10^k-1} \\
&= \frac{b}{10^N} + \frac{c}{10^N(10^k-1)}
\end{align*}

where $b = a_1a_2\cdots a_N$ and $c = a_{N+1}a_{N+2}\cdots a_{N+k}$. Clearly as the sum of two rational numbers, $x$ is rational.
\end{questionparts}