How do you find the sum of the infinite geometric series 1/2+1/4+1/8+1/16..?

1 Answer
Dec 16, 2015

Answer:

#sum_(n=1)^oo(1/2)*(1/2)^(n-1)=1#

Explanation:

Let the series be represented by #sum_(n=1)^oo(x_n)#
The common ratio of this infinite geometric series is
#r=x_(n+1)/x_n=1/4/1/2=1/2#

Since #|r|=1/2<1#, it implies that this series converges to the value #a/(1-r)#, where #a# is the first term in the series.

#therefore sum_(n=1)^oo(1/2)*(1/2)^(n-1)=a/(1-r)=(1/2)/(1-1/2)=1#.