<s>[INST] How to write $4.13\bar{9}$ in explicit form I want to write $4.13\bar{9}$ in an explicit form.
I know how to solve a problem similar to this, for a number like $2.\overline{17}$, where the bar starts at the first number after the comma:
I know that 
$$2.\overline{17}=$$
$$2+\frac{17}{100}\sum_{n=1}^\infty(\frac{1}{100})^{n-1}=$$
$$2+\frac{17}{100}\frac{1}{1-\frac{1}{100}=}$$
$$2+\frac{17}{99}=\frac{215}{99}.$$
But I'm not 100% sure how to solve a problem where the bar starts at the third number after the comma. 
Do I start the sum from $n=1$, like I did above? Do I start it for $n=3$ ?
I mean, would I write
$$4+\frac{13}{100}+\sum_{n=3}^\infty{\frac{9}{10^n}}$$or
$$4+\frac{13}{100}+\sum_{n=1}^\infty{\frac{9}{10^n}}$$
?
I can't seem to understand which one to use.
 [/INST] $4.13\bar{9} = 4  + \frac{13}{100} + \frac{9}{1000} + \frac{9}{10000} + \frac{9}{100000} + \ldots = 4 + \frac{13}{100} + \sum_{n=3}^{\infty} \frac{9}{10^n} = 4 + \frac{13}{100} + \frac{\frac{9}{1000}}{(1 - \frac{1}{10})}$. 
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