How do I use a geometric series to prove that $0.999...=1$ ?

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Answer 1 · 2014-10-04T02:46:48

$0.999...$

by separating digits,

$=0.9+0.09+0.009+cdots$

by rewriting in the form of a geometric series,

$=9/10+9/10(1/10)+9/10(1/10)^2+cdots$

by the sum formula for a convergent geometric series,

$={9/10}/{1-1/10}=1$

I hope that this was helpful.

How do I use a geometric series to prove that 0.999...=10.999...=1?