I'm having some trouble with this question: Prove that 0.1636363636...=9/55, using infinite series. I'd appreciate any help you can give me. Thanks!
Proving that a repeating decimal equals a fraction
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sequences-and-series
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0See [this](http://math.stackexchange.com/questions/29638). – 2011-05-13
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1possible duplicate of [How do we find a fraction with whose decimal expansion has a given repeating pattern?](http://math.stackexchange.com/questions/29638/how-do-we-find-a-fraction-with-whose-decimal-expansion-has-a-given-repeating-patt) – 2011-05-13
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2I did take a quick glance at that one but was a little daunted by the complexity of the explanations. – 2011-05-13
1 Answers
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You can do this: \begin{align*} 0.1636363\cdots &= \frac{1}{10} + \biggl[ 63 \times 10^{-3} + 63\times 10^{-5} + 63 \times 10^{-7} + \cdots\biggr] \\ &= \frac{1}{10} + 63 \cdot \Bigl[ 10^{-3} + 10^{-5} + 10^{-7} + \cdots \Bigr] \\ &= \frac{1}{10} + 63 \cdot \frac{10^{-3}}{1-10^{-2}} \\ &= \frac{1}{10} + \frac{7}{110}=\frac{9}{55}\end{align*}
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0I get the general idea but I'm not sure if I understand the step you took between the 3rd and 4th lines. I'm supposed to solve this problem with a series, which is the part I'm having the most trouble with. – 2011-05-13
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0@user4773: Sum of an infinte geometric series $a+ar+ar^{2} + \cdots$ is given by $\frac{a}{1-r}$. Thats what i have used and the rest are basic calculations. – 2011-05-13
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0Ahh, I see. Thank you! – 2011-05-13
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0@user4773: Welcome. – 2011-05-13
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0I removed the $\infty$; think it was a typo. – 2011-05-13
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0@Rasmus: By putting $\infty$ i actually meant that there are infinite number of terms. Anyhow thanks for the edit. It looks better without $\infty$. – 2011-05-13