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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 $(10-1)^2=81$.

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

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