# Patterns in $\frac{80}{81}$ and $\frac{10}{81}$.

The decimal form of $\frac{80}{81}$ is $0.987654320\ldots$ notice the expected $1$ is missing. The decimal form of $\frac{10}{81}$ is $0.12345679\ldots$ notice the expected $8$ is missing. Can someone expansion why the decimal form is the way they are? I think it has something to do with $(10-1)^2=81$.

• Or the way the fractions of powers of nine are represented in decimal notation. – Michael McGovern Jan 21 '17 at 4:57

This is because $\dfrac1{(1-x)^2} =\sum_{n=0}^{\infty} (n+1)x^n$.
Putting $x=.1$, this is $\dfrac1{.9^2} =\dfrac1{.81} =\dfrac{100}{81} =1+2/10+3/100+4/1000 + ...$ so $\dfrac{10}{81} =1/10+2/100+3/1000+4/10000 + ... =0.1234567...$.
The next terms are $8/10^8+9/10^9 +10/10^{10}+...$, but we get a carry here (from the $10/10^{10}$) and these terms have a value of $9/10^8+0/10^9 +0/10^{10}+...$ which explains the $....6790...$.
The other is just $1-x$ where $x$ is a decimal: $\dfrac{80}{81} =1-\dfrac{1}{81}$.