How do you find the sum of the infinite geometric series 0.9 + 0.09 + 0.009 +…?

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Answer 1 · 2016-02-13T22:39:49

It will be #0.bar(9)# which is taken as equal to 1

For instance:

#1/3=0.bar(3)#

#3xx1/3=0.bar(9)# but is also equal to 1, therefore both numbers are equal.

Answer 2 · 2016-02-13T23:23:46

Apply the geometric series formula for #a=9/10# and #r=1/10# to find that #0.9+0.09+0.009+... = 1#

The other answer is correct, but to add how applying the geometric series formula would give the result:

For #|r| < 1# and #a!=0# we have #sum_(n=0)^oor^n = a/(1-r)#

(for a short derivation of this formula, see this answer)

Applying this, as #|1/10| < 1#, we have

#0.9+0.09+0.009+... = sum_(n=0)^oo9/10(1/10)^n#

#=(9/10)/(1-1/10)#

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

#=1#