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.
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
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.
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 \(0<r<1\).
After \(N\) repeated blocks, \(S_N=\frac{B}{10^n}(1+10^{-n}+\cdots+10^{-(N-1)n})\).
With \(r=10^{-n}\), \(S_N=\frac{B}{10^n}\frac{1-r^N}{1-r}=\frac{B}{10^n}\frac{1-10^{-Nn}}{1-10^{-n}}\).
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,
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
The general partial sum is \(S_N=\sum_{k=1}^{N}27/10^{2k}\). Using the geometric sum result, its limit is
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.
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
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
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
Since \(45/9900=1/220\), combine the fractional parts using denominator \(1100\): \(13/100=143/1100\) and \(1/220=5/1100\). Therefore
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
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.
- Write the first four partial sums for \(0.\overline{4}\), then find its exact fraction.
- Convert \(0.\overline{18}\) to a fraction and reduce it to lowest terms.
- For \(0.2\overline{3}\), identify the prefix and repeating-block contributions before finding the fraction.
- Convert \(1.07\overline{2}\) to a fraction. State the values of \(A\), \(m\), \(B\), and \(n\) used.
- Explain why \(0.\overline{03}\) uses \(n=2\), even though the integer represented by its block is \(3\).
