A periodic decimal expansion sequence This is from my analysis homework.
Let $(x_n)_{n \in \mathbb{N}}$ be a sequence with the following terms.
$$x_1 = 0,1$$
$$x_2 = 0,13$$
$$x_3 = 0,131$$
$$x_4 = 0,1313$$
$$x_5 = 0,13131$$
and son on. Prove $\lim_{n \to \infty} x_n = \frac{13}{99}$.
My solution is based in two claims (easy to prove by induction). The first one is
$$ 0 \leq x_n - x_{n-2} \leq \frac{1}{10^{n-2}}$$, for $n \geq 3$. And second one is
$$100 x_n = 13 + x_{n-2}$$, for $n \geq 3$.
Now we proof using the limit definition. Consider $\varepsilon > 0$ and $n \in \mathbb{N}$, since $\lim_{n \to \infty} \frac{1}{10^n} = 0$, exists a natural number $N$, such that, $$\frac{1}{10^n} < 99 \varepsilon$$ if $n \geq N$. So, if $n \geq \text{max}\{N,3 \}$ we have
$$ 0 \leq x_n - x_{n-2} \leq \frac{1}{10^n}< 99 \varepsilon$$
$$ |x_n - x_{n-2}| < 99 \varepsilon$$
$$ |x_n - (100 x_n - 13) | < 99 \varepsilon$$
$$ |-99 x_n + 13) | < 99 \varepsilon$$
$$ |99 x_n - 13| < 99 \varepsilon$$
$$ | x_n - \frac{13}{99} | < \varepsilon$$
My question is if there exists a more simple solution because this homework is based on only limit definition, so it can be solved without using "advanced" theorems.
 A: I'll start with an independent proof that $\frac{13}{99} = 0.\overset{\mbox{.}}{1}\overset{\mbox{.}}{3} :$
Let $\ x=0.\overset{\mbox{.}}{1}\overset{\mbox{.}}{3}.$
Then $100x = 13.\overset{\mbox{.}}{1}\overset{\mbox{.}}{3}$
$99x = 100x-x = 13.\overset{\mbox{.}}{1}\overset{\mbox{.}}{3} - 0.\overset{\mbox{.}}{1}\overset{\mbox{.}}{3} = 13$
$x = \frac{13}{99} = 0.\overset{\mbox{.}}{1}\overset{\mbox{.}}{3}.$
$$$$
$\frac{13}{99} = 0.\overset{\mbox{.}}{1}\overset{\mbox{.}}{3}=0.13131313\ldots\ .$
Hence for all natural numbers $k$, we have: $\ 0<\frac{13}{99} - x_k < \frac{1}{10^k}. $
Take limits as $k\to \infty$ of the above inequality and quote the Squeeze Theorem.
An alternative method would be to prove that the sequence is Cauchy which is similar to the above method in fact, and then use the fact that every real Cauchy sequence (with the usual metric) is convergent.
A: Note that, if $n\in\Bbb N$\begin{align}x_{2n}&=\sum_{k=1}^n\frac{13}{100^k}\\&=\frac{13}{100}\sum_{k=0}^{n-1}\left(\frac1{100}\right)^k\\&=\frac{13}{100}\frac{1-\left(\frac1{100}\right)^n}{1-\frac1{100}}\\&=13\frac{1-\left(\frac1{100}\right)^n}{99}.\end{align}Can you take it from here?
