Chapter 8: Infinite Sequences and Series
Section 8.2: Series
Example 8.2.9
Write the repeating decimal 3.45&conjugate0; as the ratio of two integers.
Solution
Mathematical Solution
The following elementary approach appears to be mechanically simpler than the approach found in the series chapter of some calculus texts.
Set S=0.45&conjugate0; = 0.454545⋯, so 100 S=45.45&conjugate0; = 45.454545⋯.
Then, 100 S−S=99 S=45.0 and S=45/99 so that the given repeating decimal is 3+45/99=38/11.
The charm of this approach is that a simple algorithm can be extracted, namely, divide the repeating digits by 10k−1, where k is the number of digits that are repeated.
Calculus texts, looking for an application of the geometric series, approach the problem as follows.
0.45&conjugate0;
= 45102+45104+⋯+45102n+⋯
=451021+1102+⋯+1102n−1+⋯
=45102 11−1/102
=45100 10099
=4599
As in the simpler method, the final result is 3+45/99=38/11.
Of course, the series advocate will claim that the given repeating decimal can immediately be represented as
3+45∑n=1∞1102n
=3+4511−1/100−1
=3+451/1001−1/100
=3+4510010099
=3+4599
=34299
=3811
Maple Solution
Enter the series representation of the given number.
Context Panel: Evaluate and Display Inline
3+45∑n=1∞1102n = 3811
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