How do you find the sum of the infinite geometric series given #Sigma 40(3/5)^(n-1)# for n=1 to #oo#?

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#sum_(n=1)^oo 40(3/5)^(n-1)#

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Answer 1 · 2018-04-15T02:16:38

#100#

The sum of an infinite geometric series in the below form is given by

#sum_(n=1)^ooa(r)^(n-1)=a/(1-r),# so long as #-1<r<1.#

Here, for #sum_(n=1)^oo40(3/5)^(n-1), |r|=3/5<1#, so it will have a finite sum. Also #a=40,# so

#sum_(n=1)^oo40(3/5)^(n-1)=40/(1-3/5)=40/(2/5)=40*5/2=100#