I have to show a repeating decimal sequence converges. For example I have a sequence $\{x_n\}$ where $x_n = 0.121212...1212...$
I know that this decimal equals $4/33$. I try to find information online, but everything shows up says to make it as a geomtric series and proving it as that. However I don't think that is correct since in a sense sequences and series are different.
Any help or references for proving this type of sequence?
On
Starting from the fact that $0.9999999\ldots=0.\overline{9}=1$, we have $$0.\overline{1}=\frac{1}{9},\quad 0.\overline{01}=\frac{1}{99},\quad 0.\overline{001}=\frac{1}{999},\quad 0.\overline{0001}=\frac{1}{9999}$$ and so on. It follows that any number whose decimal representation is purely periodic is an integer multiple of $\frac{1}{10^k-1}$ for some $k$. In your case $0.\overline{12} = 12\cdot 0.\overline{01} = \frac{12}{99} = \frac{4}{33}.$
It's a geometric series, times 12: $$ 0.12 = \frac{12}{100^1},\quad 0.0012 = \frac{12}{100^2},\quad 0.000012 = \frac{12}{100^3},\quad\cdots. $$ Therefore $$ \begin{aligned} x_n &= \frac{12}{100^1} + \frac{12}{100^2} + \frac{12}{100^3} + \cdots\\ &= 12\bigg(\frac{1}{100^1}+\frac{1}{100^2}+\frac{1}{100^3}+\cdots\bigg) \\ & = 12\sum_{n=1}^\infty \frac1{100^n}\\ &= 12 \bigg(\frac{\frac1{100}}{1 - \frac1{100}}\bigg)\\ &= \frac{12}{99}\\ &= \frac4{33} \end{aligned} $$