Express the repeating decimal $0.090909\ldots$ as a quotient of two integers.

Solution We write the infinite decimal $0.090909\ldots$ as an infinite series: $$\begin{eqnarray*} 0.090909\ldots &=& 0.09 + 0.0009+0.000009+0.00000009+\cdots \\ &=& \dfrac{9}{100}+\dfrac{9}{10000} +\dfrac{9}{1000000}+\cdots \\ &=& \dfrac{9}{100}\left( 1+\dfrac{1}{100}+\dfrac{1}{10000}+\dfrac{1}{1000000} +\cdots\right) \\ &=& \sum\limits_{k\,=\,1}^{\infty }\dfrac{9}{100}\left( \frac{1}{ 100}\right) ^{k-1} \end{eqnarray*}$$

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This is a geometric series with $a=\dfrac{9}{100}$ and $r=\dfrac{1}{100}$. Since $\left\vert  r\right\vert <1$, the series converges and its sum is $$\sum\limits_{k\,=\,1}^{\infty }\dfrac{9}{100}\left( \frac{1}{100}\right) ^{k-1}=\frac{\dfrac{9}{100}}{1-\dfrac{1}{100}}=\frac{9}{99}=\frac{1}{11}$$

So, $0.090909\ldots \,=\dfrac{1}{11}$.