# How to prove a repeating decimal sequence converges?

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?

• All decimal sequences converge, and for the same reason. – Gae. S. Sep 10 '19 at 22:24
• The idea is that the repeating decimal sequence is equivalent to the partial sums of a geometric series. So, showing that the series converges is equivalent to showing the decimal sequence converges. – D.B. Sep 10 '19 at 22:27
• Also, sequences and series are not all that different. Series are just "special" sequences (technically, every sequence may also be written as a series), where the general term is $s_n=\sum_{k=1}^n a_k$ – Gae. S. Sep 10 '19 at 22:28
• You will inevitably talk about series when you talk about decimal sequences, because they represent a series. It's how they're defined: .324 means $3\cdot 10^{-1} + 2 \cdot 10^{-2} + 4 \cdot 10^{-3}$. – Jair Taylor Sep 10 '19 at 22:57

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}
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 $$0.\overline{12} = 12\cdot 0.\overline{01} = \frac{12}{99} = \frac{4}{33}.$$