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There is a common argument as to whether
0.999
…
=
1
{\displaystyle 0.999\ldots =1}
or not. It does. Convergence is the primary reason as to why.
Proof
0.999
…
{\displaystyle 0.999\ldots }
=
9
×
0.111
…
{\displaystyle =9\times 0.111\ldots \,\!}
=
9
×
(
1
10
+
1
100
+
1
1000
+
…
)
{\displaystyle =9\times \left({\frac {1}{10}}+{\frac {1}{100}}+{\frac {1}{1000}}+\ldots \right)\,\!}
=
9
×
(
−
1
+
1
1
+
1
10
+
1
100
+
1
1000
+
…
)
{\displaystyle =9\times \left(-1+{\frac {1}{1}}+{\frac {1}{10}}+{\frac {1}{100}}+{\frac {1}{1000}}+\ldots \right)\,\!}
=
9
×
(
−
1
+
∑
i
=
0
∞
(
1
10
)
i
)
{\displaystyle =9\times \left(-1+\sum _{i=0}^{\infty }\left({\frac {1}{10}}\right)^{i}\right)\,\!}
=
9
×
(
−
1
+
1
1
−
1
10
)
{\displaystyle =9\times \left(-1+{\frac {1}{1-{\frac {1}{10}}}}\right)}
=
9
×
(
−
1
+
10
9
)
{\displaystyle =9\times \left(-1+{\frac {10}{9}}\right)}
=
1
{\displaystyle =1\,}
Explaination
The key step to understand here is that
1
1
+
1
10
+
1
100
+
1
1000
+
…
=
∑
i
=
0
∞
(
1
10
)
i
{\displaystyle {\frac {1}{1}}+{\frac {1}{10}}+{\frac {1}{100}}+{\frac {1}{1000}}+\ldots =\sum _{i=0}^{\infty }\left({\frac {1}{10}}\right)^{i}\,\!}
Template:Mathstub
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