What is that geometric series thing he was talking about?

professionalbronco

What is that geometric series thing he was talking about? (I know the way that AoPS teaches, but I never thought of a way with geometric series.)

quacker88

@professionalbronco sorry for getting to you so late!

here's the idea with a geometric series:
say we have some repeating decimal, let's use $0.\overline{81}$.
If we break it down into its repeating parts, it might look something like this:
$$0.\overline{81}=0.810000+0.008100+0.000081+...$$ so on and so forth.
Each term has the 81 shifted over by two decimal places, so in other words, each term is $\frac{1}{100}$ of the previous term.

From here, there's two ways to tackle an infinite geometric series: one way is to work it out algebraically.
Let's call the infinite sum S, then:
$S=\frac{81}{100}+\frac{81}{10000}+\frac{81}{1000000}+...$
Since each term is $\frac{1}{100}$ of the previous term, we can manipulate the equation by multiplying S by $\frac{1}{100}$.

$\frac{S}{100}=\frac{81}{10000}+\frac{81}{1000000}+\frac{81}{100000000}+...$
wait... so every term in $\frac{S}{100}$ is going to match up with every term in $S$, except for the initial $\frac{81}{100}$. So, if we take $S–\frac{S}{100}$, everything cancels out and we get

$S–\frac{S}{100} = \frac{81}{100} \implies \frac{99S}{100}=\frac{81}{100} \implies \boxed{S=\frac{81}{99}=\frac{9}{11}}$

The other method is just plugging into the formula for a geometric series, where $a$ is the first term and $r$ is the common ratio: $\frac{a}{1-r}$

This formula actually comes just from using the method in the first part, just with variables instead. Give it a shot!

Anyways, plugging $a=\frac{81}{100}$ and $r=\frac{1}{100}$ we get $\frac{\frac{81}{100}}{1-\frac{1}{100}}\implies \frac{\frac{81}{100}}{\frac{99}{100}}\implies \frac{81}{99}$ as with the first method.

Also, just using the patterns with repeating decimals, since $0.\overline{81}$ repeats with a period of 2 (it repeats every 2 digits), we know its denominator is $99$.

Bolun Zhang

Thank you