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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!

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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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  • $\begingroup$ I 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. $\endgroup$ – Matthew May 13 '11 at 7:18
  • $\begingroup$ @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. $\endgroup$ – user9413 May 13 '11 at 7:22
  • $\begingroup$ Ahh, I see. Thank you! $\endgroup$ – Matthew May 13 '11 at 7:25
  • $\begingroup$ @user4773: Welcome. $\endgroup$ – user9413 May 13 '11 at 7:27
  • $\begingroup$ I removed the $\infty$; think it was a typo. $\endgroup$ – Rasmus May 13 '11 at 8:07

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