How do you find the sum of the infinite geometric series #Sigma 2(2/3)^n# from n=0 to #oo#?

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Answer 1 · 2018-04-27T04:01:15

#sum2(2/3)^n = 6#

#sum2(2/3)^n# for n=0 to infinity

=2#(sum(2/3)^n)#

Sum of an infinite geometric series (where the ratio r is #-1<=r<=1#) is given by #a/(1-r)# where r is the common ratio and ‘a’ is the first term.

Therefore,
#sum(2/3)^n = 1/(1-(2/3))#

#=1/(1/3) = 3#

Substituting this value in our original equation gives us -

2#(sum(2/3)^n) = 2*3 = 6#