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Series and Partial Sums · Tutorial 562 of 998

Repeating Decimals as Geometric Series

Convert repeating decimal digits into geometric-series terms and use the limiting partial sums to derive and verify exact fractional values.

Beginner 9 min read

What You'll Learn

  • Interpret a repeating block as a sequence of decimal contributions.
  • Derive the fraction for a pure repeating decimal from its partial sums.
  • Handle a finite nonrepeating prefix before the repeating block.
  • Apply a general formula while keeping track of decimal-place shifts.
  • Check exact fraction results and avoid treating a truncation as the full decimal.

Why Repeating Decimals Lead to Series

A repeating decimal is not merely a finite string of digits with a bar over some of them. It represents an infinite sum of decimal-place contributions. Recognizing those contributions as a geometric series gives a direct way to find the exact fraction represented by the decimal.

For example, the decimal \(0.\overline{3}\) means that the digit 3 continues forever after the decimal point. Its successive finite truncations are \(0.3\), \(0.33\), \(0.333\), and so on. Each new truncation adds a smaller contribution, and the truncations approach the value of the infinite decimal.

Definition: In \(0.\overline{27}\), the bar indicates that the block of digits \(27\) repeats indefinitely: \(0.272727\ldots\). A finite decimal truncation contains only finitely many of those repeating digits.

The key observation is that each repeated block contributes the same digits shifted a fixed number of decimal places farther to the right. Those contributions therefore have a constant ratio, which is exactly the structure of a geometric series.

Pure Repeating Decimals

A pure repeating decimal has no nonrepeating digits after the decimal point before its repeating block begins. Suppose a block of \(n\) digits represents the integer \(B\). For instance, the block \(27\) has \(n=2\) and \(B=27\). Its first contribution is \(B/10^n\), the next is \(B/10^{2n}\), and each subsequent contribution is obtained by multiplying by \(10^{-n}\).

After \(N\) copies of the block, the decimal truncation is the finite sum

$$ S_N=\frac{B}{10^n}+\frac{B}{10^{2n}}+\cdots+\frac{B}{10^{Nn}} =\sum_{k=1}^{N}\frac{B}{10^{kn}}. $$

This is a geometric sum with first term \(B/10^n\) and common ratio \(10^{-n}\). Since \(0<10^{-n}<1\), its partial sums approach a limit. The finite geometric sum formula gives the exact partial sum and, on taking the limit, the value of the repeating decimal.

Theorem: If \(B\) is the integer represented by a repeating block of \(n\) digits, then $$ 0.\overline{B}=\frac{B}{10^n-1}. $$ Here \(0.\overline{B}\) means the decimal with that \(n\)-digit block repeating. If the block begins with zeroes, use its full length for \(n\); for example, \(0.\overline{03}\) has \(B=3\) and \(n=2\).

Why the Pure-Repeat Formula Works

For a ratio \(r\) with \(r\ne 1\), the first \(N\) powers satisfy \(1+r+\cdots+r^{N-1}=(1-r^N)/(1-r)\). In the repeating-decimal sum, the ratio is \(r=10^{-n}\). Multiplying the finite sum by \(1-r\) cancels all but the first and last terms. The remaining last term tends to zero because \(0<r<1\).

1
Write the finite truncation.
After \(N\) repeated blocks, \(S_N=\frac{B}{10^n}(1+10^{-n}+\cdots+10^{-(N-1)n})\).
2
Apply the finite geometric sum formula.
With \(r=10^{-n}\), \(S_N=\frac{B}{10^n}\frac{1-r^N}{1-r}=\frac{B}{10^n}\frac{1-10^{-Nn}}{1-10^{-n}}\).
3
Take the limit of the truncations.
As \(N\) increases, \(10^{-Nn}\) tends to zero. Thus the decimal’s value is \(\frac{B}{10^n}\frac{1}{1-10^{-n}}=\frac{B}{10^n-1}\).

This derivation connects the decimal to its partial sums: the infinite decimal’s value is the limit of the finite truncations. It also explains why the denominator contains a string of \(n\) nines, since \(10^n-1\) is that number.

Worked Example: Convert \(0.\overline{3}\) to a Fraction

The repeating block is \(3\), so \(B=3\) and \(n=1\). The first few partial sums are \(0.3\), \(0.3+0.03=0.33\), and \(0.3+0.03+0.003=0.333\). In general,

$$ S_N=\frac{3}{10}+\frac{3}{10^2}+\cdots+\frac{3}{10^N} =\frac{3}{10}\frac{1-10^{-N}}{1-10^{-1}}. $$

Since \(1-10^{-1}=9/10\), this becomes \(S_N=\frac{3}{9}(1-10^{-N})\). As \(N\) tends to infinity, \(10^{-N}\) tends to zero, so the decimal’s value is \(3/9=1/3\).

Check: \(3(1/3)=1\), and the long-division digits for \(1/3\) are \(0.333\ldots\). Thus the fraction has the stated repeating decimal expansion.

Worked Example: Convert \(0.\overline{27}\) to a Fraction

The block \(27\) has length \(n=2\), so each new copy is shifted two decimal places to the right. The first three partial sums are

$$ S_1=0.27,\qquad S_2=0.27+0.0027=0.2727,\qquad S_3=0.27+0.0027+0.000027=0.272727. $$

The general partial sum is \(S_N=\sum_{k=1}^{N}27/10^{2k}\). Using the geometric sum result, its limit is

$$ \frac{27}{10^2-1}=\frac{27}{99}=\frac{3}{11}. $$

Check: \(99(3/11)=9\cdot3=27\), so the reduction from \(27/99\) is correct. Long division confirms the repeating digits: \(3\div11=0\) remainder \(3\); \(30\div11=2\) remainder \(8\); \(80\div11=7\) remainder \(3\). The same remainders recur, so the digits \(27\) repeat.

When a Nonrepeating Prefix Comes First

A mixed repeating decimal has a finite prefix of nonrepeating decimal digits followed by a repeating block. For example, \(0.1\overline{6}\) has one nonrepeating digit, \(1\), and then the digit \(6\) repeats. Separate the finite prefix from the contributions of the repeated block. The latter still form a geometric series, but their first term must be shifted to the correct decimal place.

Formula: Let \(A\) be the integer made from the \(m\) nonrepeating decimal digits, and let \(B\) be the integer made from the \(n\) repeating digits. Then $$ 0.\text{(prefix)}\overline{\text{(block)}} =\frac{A}{10^m}+\frac{B}{10^m(10^n-1)}. $$ If there is no nonrepeating prefix, take \(m=0\) and \(A=0\). The repeated block’s first contribution is \(B/10^{m+n}\), and each later one is multiplied by \(10^{-n}\).

To see the placement in the formula, the repeated part begins after \(m\) prefix digits. Its contributions are \(B/10^{m+n}\), \(B/10^{m+2n}\), and so on. The first term of this geometric series is \(B/10^{m+n}\), and its ratio is \(10^{-n}\). Its sum is therefore

$$ \frac{B}{10^{m+n}}\frac{1}{1-10^{-n}} =\frac{B}{10^m(10^n-1)}. $$

The prefix contributes \(A/10^m\), so adding the prefix and the repeated part gives the formula.

Worked Example: Convert \(0.1\overline{6}\) to a Fraction

The nonrepeating prefix is \(1\), so \(A=1\) and \(m=1\). The repeating block is \(6\), so \(B=6\) and \(n=1\). The decimal is the prefix \(1/10\) plus the repeating contributions \(6/100+6/1000+6/10000+\cdots\).

The repeated part has first term \(6/100\) and ratio \(1/10\). Its sum is

$$ \frac{6}{100}\frac{1}{1-\frac{1}{10}} =\frac{6}{100}\frac{10}{9} =\frac{60}{900} =\frac{1}{15}. $$

Now add the prefix: \(1/10+1/15=3/30+2/30=5/30=1/6\). Therefore \(0.1\overline{6}=1/6\).

Check: dividing \(1\) by \(6\) gives \(0\) remainder \(1\); then \(10\div6=1\) remainder \(4\); then \(40\div6=6\) remainder \(4\). The remainder \(4\) repeats, producing \(0.1666\ldots\), as required.

Worked Example: Convert \(2.13\overline{45}\) to a Fraction

The integer part is \(2\). The two nonrepeating decimal digits form \(A=13\), with \(m=2\), and the repeating block is \(B=45\), with \(n=2\). Thus the number is

$$ 2+\frac{13}{100}+\frac{45}{100(10^2-1)} =2+\frac{13}{100}+\frac{45}{9900}. $$

Since \(45/9900=1/220\), combine the fractional parts using denominator \(1100\): \(13/100=143/1100\) and \(1/220=5/1100\). Therefore

$$ 2+\frac{143}{1100}+\frac{5}{1100} =2+\frac{148}{1100} =2+\frac{37}{275} =\frac{587}{275}. $$

Check the calculation in reverse. The prefix and repeated block give \((21345-213)/(100\cdot99)=21132/9900\). Dividing numerator and denominator by \(36\) gives \(587/275\), the same fraction. The first digits are \(2.13\), followed by the block \(45\) repeating, as in the original decimal.

A Common Mistake: Stopping at a Truncation

A finite truncation is only a partial sum, not the value of the infinite repeating decimal. For example, one might incorrectly write

$$ 0.1\overline{6}=0.1+0.06=0.16=\frac{4}{25}. $$

This calculation stops after including just one of the repeated \(6\) digits. The actual decimal also includes \(0.006+0.0006+\cdots\). Those omitted terms have a positive sum, so \(0.16\) cannot be the value of the full decimal. Their sum is \(0.06(1/10)/(1-1/10)=1/150\), and \(0.16+1/150=24/150+1/150=25/150=1/6\).

Another useful check is to substitute the resulting fraction into the appropriate decimal formula. A correct simplification should preserve the value; a fraction that gives only the first few digits is not yet an exact conversion.

Check Your Understanding

Use partial sums and the geometric-series formulas to answer each question. Keep track of how many decimal places precede the repeating block.

  1. Write the first four partial sums for \(0.\overline{4}\), then find its exact fraction.
  2. Convert \(0.\overline{18}\) to a fraction and reduce it to lowest terms.
  3. For \(0.2\overline{3}\), identify the prefix and repeating-block contributions before finding the fraction.
  4. Convert \(1.07\overline{2}\) to a fraction. State the values of \(A\), \(m\), \(B\), and \(n\) used.
  5. Explain why \(0.\overline{03}\) uses \(n=2\), even though the integer represented by its block is \(3\).

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