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Revision as of 19:48, 6 May 2005 editBradBeattie (talk | contribs)6,888 editsm →Explanation← Previous edit Revision as of 10:56, 7 May 2005 edit undo68.198.149.91 (talk) substituted k for i to avoid confusion with imaginary unitNext edit →
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|<math>= -9 + 9 \times \sum_{i=0}^\infty \left( \frac{1}{10} \right)^i</math> |<math>= -9 + 9 \times \sum_{k=0}^\infty \left( \frac{1}{10} \right)^k</math>
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The key step to understand here is that The key step to understand here is that
:<math>\sum_{i=0}^\infty \left( \frac{1}{10} \right)^i = \frac{1}{1 - \frac{1}{10}}.</math> :<math>\sum_{k=0}^\infty \left( \frac{1}{10} \right)^k = \frac{1}{1 - \frac{1}{10}}.</math>


This is the sum of the convergent geometric series. For further information, see ] and ]. This is the sum of the convergent geometric series. For further information, see ] and ].

Revision as of 10:56, 7 May 2005

In mathematics, one could easily fall in the trap of thinking that while 0.999... is certainly close to 1, nevertheless the two are not equal. Here's a proof that they actually are.

Proof

0.999 … {\displaystyle 0.999\ldots } = 9 10 + 9 100 + 9 1000 + ⋯ {\displaystyle ={\frac {9}{10}}+{\frac {9}{100}}+{\frac {9}{1000}}+\cdots }
= − 9 + 9 1 + 9 10 + 9 100 + 9 1000 + ⋯ {\displaystyle =-9+{\frac {9}{1}}+{\frac {9}{10}}+{\frac {9}{100}}+{\frac {9}{1000}}+\cdots }
= − 9 + 9 × ∑ k = 0 ∞ ( 1 10 ) k {\displaystyle =-9+9\times \sum _{k=0}^{\infty }\left({\frac {1}{10}}\right)^{k}}
= − 9 + 9 × 1 1 − 1 10 {\displaystyle =-9+9\times {\frac {1}{1-{\frac {1}{10}}}}}
= 1. {\displaystyle =1.\,}

Explanation

The key step to understand here is that

∑ k = 0 ∞ ( 1 10 ) k = 1 1 − 1 10 . {\displaystyle \sum _{k=0}^{\infty }\left({\frac {1}{10}}\right)^{k}={\frac {1}{1-{\frac {1}{10}}}}.}

This is the sum of the convergent geometric series. For further information, see geometric series and convergent series.

External proofs


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