How do you use a geometric series to prove that #0.999…=1#?

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How do you use a geometric series to prove that #0.999…=1#?

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Answer 1 · 2014-09-12T17:11:18

We can write
#0.999...=9/10+9/100+9/1000+cdots#
#=9/10+9/10(1/10)+9/10(1/10)^2+cdots#
#=sum_{n=0}^infty 9/10(1/10)^n#,
which is a geometric series with
#a=9/10# and #r=1/10#.

So, the sum is
#a/{1-r}={9/10}/{1-1/10}=1#

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How do you use a geometric series to prove that #0.999…=1#?

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