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Revision as of 19:22, 6 May 2005 editBradBeattie (talk | contribs)6,888 edits Proof header← Previous edit Revision as of 19:43, 6 May 2005 edit undoMichael Hardy (talk | contribs)Administrators210,264 edits The value of the article titled convergence is mainly in its "see also" list; it borders on being a disambiguation page. I'm changing a link to that article to convergent series.Next edit →
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:<math>\sum_{i=0}^\infty \left( \frac{1}{10} \right)^i = \frac{1}{1 - \frac{1}{10}}.</math> :<math>\sum_{i=0}^\infty \left( \frac{1}{10} \right)^i = \frac{1}{1 - \frac{1}{10}}.</math>


For further information, see ] and ]. For further information, see ] and ].


== External proofs == == External proofs ==

Revision as of 19:43, 6 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 × ∑ i = 0 ∞ ( 1 10 ) i {\displaystyle =-9+9\times \sum _{i=0}^{\infty }\left({\frac {1}{10}}\right)^{i}}
= − 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

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

For further information, see geometric series and convergent series.

External proofs


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