Reciprocal of 9's

Question

I happened upon an interesting pattern today. If I take the reciprocal of $998$, I get $.001002004008016032064128\dots$ which has a pattern of powers of $2$. If I take the reciprocal of $997$, I get a pattern with powers of $3$, $.001003009027081\dots$ This pattern continued with $996$ having powers of $4$ and so on. I can also control how many zeros are found between the powers by adding additional nines.

My question is this: Does this pattern have a name and what is a practical application of this pattern?

Answer 1

Here is a proof of this fact, using the fact that an infinite geometric series, when convergent, has a limit that can be computed (see Remark below); for example:

$$\dfrac{1}{998}=\dfrac{1}{1000-2}=\dfrac{1}{1000}\left(\dfrac{1}{1-\tfrac{2}{1000}}\right)=$$

$$\tag{1}=\tfrac{1}{1000}\left(1+\tfrac{2}{1000}+(\tfrac{2}{1000})^2+(\tfrac{2}{1000})^3+\cdots\right)$$

$$=\tfrac{1}{1000}\left(1+\tfrac{2}{1000}+\tfrac{4}{1000000}+\tfrac{8}{1000000000}+\tfrac{16}{1000000000000}+\cdots\right)$$

$$=\tfrac{1}{1000}\left(1.0020040080160\cdots\right)$$

$$=0.0010020040080160\cdots$$

Remark: in (1) we have applied formula $1+a+a^2+a^3+\cdots = \dfrac{1}{1-a}.$

Answer 2

$$\begin{align} \frac1{998}&=\frac1{1000-2}\\ &=\frac1{1000}\frac1{1-2/1000}\\ &=\frac1{1000}\left(1+\frac{2}{1000}+\frac{2^2}{1000^2}+\frac{2^3}{1000^3}+\cdots\right)\\ &=\frac1{1000}+\frac{2}{1000^2}+\frac{2^2}{1000^3}+\frac{2^3}{1000^4}+\cdots \end{align} $$

It's the jolly old geometric series again!