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D

**Theorem** \[0.(9) = 0.9999 \ldots = 1\]

**Proof**

It is clear that \[0.(9) = 0.9999 \ldots = 0.9 + 0.09 + 0.009 + \cdots = \sum_{i=1}^{\infty}0.9\frac{1}{10^{i-1}}\] Hence $0.(9)$ is a sum of terms of infinite geometric sequence \[\frac{9}{10}, \frac{9}{10^{2}}, \frac{9}{10^{3}}, \cdots\] with the initial value equal to $\frac{9}{10}$ and common ratio $\frac{1}{10}$. Because sum of terms of infinite geometric sequence with the initial value $a$ and common ratio $|r| < 1$ is given by the following formula \[\sum_{i=1}^{\infty} ar^{i-1} = \frac{a}{1-r}\] therefore \[0.(9) = \sum_{i=1}^{\infty}0.9\frac{1}{10^{i-1}} = \frac{\frac{9}{10}}{1-\frac{1}{10}} = \frac{\frac{9}{10}}{\frac{9}{10}} = 1\] Q.E.D.

© 2007-2014, Mikołaj Hajduk