How do you find the sum of the infinite geometric series given #Sigma (3/8)(3/4)^(n-1)# from n=1 to #oo#?

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Answer 1 · 2017-02-06T08:40:19

#S_(oo)=3/2#

For a GP #" "a+ar+ar^2+ar^3+....#

#"providing " |r|<1#

#" the series will have a sum to infinity"#
#" and is calculated by the formula"#

#S_(oo)=a/(1-r)#

in this case

#sum_0^(oo)(3/8)(3/4)^(n-1)=(3/8)sum_0^(oo)(3/4)^(n-1)#

if we write out the series for #n= 1,2,3,4,....#

#S_oo=(3/8)color(blue)({1+3/4+(3/4)^2+(3/4)^3+.......})#

The GP is coloured blue.

#a=1, r=3/4#

#because |3/4|<1 " the sum to infinity exists"#

#S_(oo)=(3/8)(1/(1-(3/4)))#

#S_(oo)=(3/8)(1/(1/4))#

#S_(oo)=(3/8)xx(4/1)=12/8=3/2#