How do you find the sum of the infinite geometric series 1/8-1/16+1/32-1/64+...?

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Answer 1 · 2018-04-28T08:21:00

#S_oo=1/12#

#1/8-1/16+1/32-1/64--(1)

a GP is in the form

#a+ar+ar^2+..ar^(n-1)+..#

for #(1)#

we have #a=1/8#

to find the common ratio #r# we divide adjacent terms

#r=(-1/16)/(1/8)=-1/2#

for a GP to have a sum to infinity

#|r|<1#

#|r|=|-1/2|<1#

#:.EES_oo#

#S_oo=a/(1-r)#

#S_oo=(1/8)/(1- -1/2)#

#S_oo=(1/8)/((3/2))=1/8-:3/2#

#S_oo=1/12#