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
Nov 16, 2015
#sum_(n=0)^oo (x^(3n)-x^(3n+1))=1/(1+x+x^2)#
with radius of convergence
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
