Question

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

Answer 1

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`

Answer 2

`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`