Is this a correct proof of a fact about rational numbers? 
Prove the following statement: if $\frac{p}{q} (p \in \mathbb{N}, q \in \mathbb{N})$ is a rational number that corresponds to an infinite periodic decimal fraction $\alpha$, then the rational number $10^{k} \frac{p}{q} (k \in \mathbb{N})$ corresponds to an infinite periodic decimal fraction obtained from the decimal fraction $\alpha$ by moving the decimal separator (period) by $k$ places to the right.

Here is my proof (not sure about its correctness):

As the problem states, $\frac{p}{q} = \alpha = a_1a_2\ldots a_n.a_{n+1}a_{n+2}\ldots a_{n+k}a_{n+k+1}\ldots$ (i.e. $a_1a_2\ldots a_n$ - integer part, $a_{n+1}a_{n+2}\ldots a_{n+k}\ldots$ - fractional part), $a_i$ - decimal digit for $\forall i \in \mathbb{N}$.
$\begin{align*}
10^{k} \frac{p}{q} 
&= 10^k\alpha = 10^k(a_1a_2\ldots a_n + \frac{a_{n+1}}{10} + \frac{a_{n+2}}{10^2} + \ldots + \frac{a_{n+k}}{10^k} + \frac{a_{n+k+1}}{10^{k+1}} + \ldots) \\
&= a_1a_2\ldots a_n\underbrace{00\ldots 0}_k + a_{n+1}\underbrace{00\ldots 0}_{k-1} + a_{n+2}\underbrace{00\ldots 0}_{k-2} + \ldots \\
&\ + a_{n+k} + \frac{a_{n+k+1}}{10} + \frac{a_{n+k+2}}{10^2} + \ldots \\
&= a_1a_2\ldots a_{n+k-1}a_{n+k}.a_{n+k+1}a_{n+k+2}\ldots
\end{align*}$

The question is: Is it correct to represent an infinite decimal fraction as a sum of infinite number of rational numbers (i.e. is it always true that, for example, $0.a_1a_2\ldots a_na_{n+1}\ldots = \frac{a_1}{10} + \frac{a_2}{10^2} + \ldots + \frac{a_n}{10^n} + \frac{a_{n+1}}{10^{n+1}} + \ldots$?)
 A: This really depends on what level we are talking about.
Formally, the definition of the decimal expansion is that it is equal to the limit of the infinite sum/series. That is, by definition, when we write
$$0.a_1a_2a_3\cdots a_n\cdots$$
we really mean
$$\sum_{i=1}^{\infty}\frac{a_i}{10^i},$$
by which we really mean
$$\lim_{n\to\infty}\left(\sum_{i=1}^n\frac{a_i}{10^i}\right).$$
So not only is it correct, it's what it is. But this requires you to know what limits and series are.
If you do, that gives you a way of writing out what you did a bit clearer. If you have
$$\frac{p}{q} = \sum_{i=1}^{\infty}\frac{a_i}{10^i},$$
then
\begin{align*}
10^k\frac{p}{q} &= 10^k\sum_{i=1}^{\infty}\frac{a_i}{10^i}\\
&= 10^k\lim_{n\to\infty}\left(\sum_{i=1}^{n}\frac{a_i}{10^i}\right)\\
&= \lim_{n\to\infty}\left(\sum_{i=1}^n\frac{10^ka_i}{10^i}\right)\\
&= \lim_{n\to\infty}\left(10^{k-1}a_1 + 10^{k-2}a_2+\cdots+a_k + \sum_{i=k+1}^{n}\frac{a_i}{10^{i-k-1}}\right)\\
&= 10^{k-1}a_1 + 10^{k-2}a_2 + \cdots + a_k + \lim_{n\to\infty}\sum_{i=k+1}^n\frac{a_i}{10^{i-k-1}}\\
&= 10^{k-1}a_1 + 10^{k-2}a_2+\cdots + a_k + \lim_{n\to\infty}\sum_{j=1}^{n-k-1}\frac{a_{j+k}}{10^{j}}\\
&= 10^{k-1}a_1+10^{k-2}a_2+\cdots+a_k + \sum_{j=1}^{\infty}\frac{a_{j+k}}{10^{j}}.
\end{align*}
which when translated back into decimal notation is
$$a_1a_2\cdots a_k.a_{k+1}a_{k+2}\cdots$$
