Recurring Decimal as a Geometric Series

A recurring decimal is a geometric series in disguise. Split it into repeating blocks, identify $a$ and $r$, and use the sum to infinity $S_\infty = \frac{a}{1-r}$ to get the exact fraction.

Express the recurring decimal $0.\dot{4}\dot{7} = 0.474747\ldots$ as a fraction in its lowest terms, using the sum to infinity of a geometric series. [4]

The working

Step 1, write it as a sum of repeating blocks: $0.474747\ldots = 0.47 + 0.0047 + 0.000047 + \cdots \quad \text{(M1)}$

Step 2, identify $a$ and $r$. Each block is $\frac{1}{100}$ of the one before: $a = 0.47, \qquad r = 0.01 \quad \text{(A1)}$

Since $|r| = 0.01 < 1$, the sum to infinity converges.

Step 3, apply $S_\infty = \dfrac{a}{1-r}$: $S_\infty = \frac{0.47}{1 - 0.01} = \frac{0.47}{0.99} \quad \text{(M1)}$

Step 4, clear the decimals to a fraction in lowest terms: $= \frac{47}{99} \quad \text{(A1)}$

Since $47$ is prime and does not divide $99$, this is already in lowest terms.

Where the marks are won and lost

• The common ratio matches the block length: a two-digit repeat gives $r = 0.01$. A three-digit repeat (e.g. $0.\dot{1}2\dot{3}$) would give $r = 0.001$.
• Multiply numerator and denominator by $100$ to turn $\frac{0.47}{0.99}$ into $\frac{47}{99}$.
• Check the fraction is fully simplified before finishing.

Common mistakes

• Taking $r = 0.1$ instead of $0.01$ for a two-digit block.
• Using $a = 0.4$ rather than the full block $0.47$.
• Leaving the answer as $\frac{0.47}{0.99}$ instead of a proper fraction.

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