How do you find the sum of the infinite geometric series 0.5 + 0.05 + 0.005 + ...?

2 Answers
Konstantinos Michailidis
Feb 21, 2016

We notice that

#0.5 + 0.05 + 0.005 + ...=5*[1/10+1/100+1/1000+...]= 5*[1/10+1/10^2+1/10^3+...]#

But #[1/10+1/10^2+1/10^3+...]# it is a geometric progression with
first term #a_1=1/10# and ratio #r=1/10#

hence its sum is given by

#S=a_1*1/(1-r)=1/10*1/(1-1/10)=1/9#

So the initial sum is #5/9#

Jim G.
Feb 21, 2016

#S_∞ = 5/9 #

Explanation:

The sum to n terms of a geometric sequence is

#S_n =( a( 1 - r^n ))/(1 - r) #

As # n → ∞ " then " r^n → 0 #

and # S_∞ = a/(1 - r )# [ for -1 < r < 1 ]

where a , is the first term and r , the common ratio

here a = #0.5 = 1/2 " and " r = 0.5/0.5 = 0.005/0.05 = 1/10 #

#rArr S_∞ =( 1/2)/(1 - 1/10) =( 1/2)/(9/10) = 1/2xx10/9 = 5/9 #

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