?How do you find the sum of the infinite geometric series 0.03, 0.03, 0.003?

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Answer 1 · 2015-12-16T04:22:22

#0.3 + 0.03 + 0.003 + ... = 1/3#

The sum of a geometric series with initial term #a# and common ratio #r# where #|r|<1# is given by

#sum_(n=0)^ooar^n = a/(1-r)#

(A derivation of this result is included in this answer)

Applying this to the given series, we get

#0.3 + 0.03 + 0.003 + ... = sum_(n=0)^oo3/10(1/10)^n#

#= (3/10)/(1-1/10)#

#= (3/10)/(9/10)#

#=1/3#

Note that this result intuitively makes sense just looking at the series, as

#0.3 + 0.03 + 0.003 + ... = 0.bar3 = 1/3#