How do you find the sum of the infinite geometric series #(1 - x) + (x^3 – x^4) + (x^6 – x^7) + ….. + [x^(3n) - x^(3n+1)] #?

1 Answer
George C.
Nov 16, 2015

#sum_(n=0)^oo (x^(3n)-x^(3n+1))=1/(1+x+x^2)#

with radius of convergence #1#

Explanation:

#sum_(n=0)^oo (x^(3n)-x^(3n+1))#

#=(1-x) sum_(n=0)^oo x^(3n)#

#=(1-x) sum_(n=0)^oo (x^3)^n#

#=(1-x)/(1-x^3)(1-x^3)sum_(n=0)^oo (x^3)^n#

#=(1-x)/(1-x^3)(sum_(n=0)^oo (x^3)^n - x^3 sum_(n=0)^oo (x^3)^n)#

#=(1-x)/(1-x^3)(sum_(n=0)^oo (x^3)^n - sum_(n=1)^oo (x^3)^n)#

#=(1-x)/(1-x^3)#

#=(1-x)/((1-x)(1+x+x^2))#

#=1/(1+x+x^2)#

with radius of convergence #1#

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