How do you find the interval of convergence #Sigma 3^nx^(2n)# from #n=[0,oo)#?

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Answer 1 · 2017-04-24T21:02:57

#sum_(n=0)^oo 3^nx^(2n) = 1/(1-3x^2)# for #absx < 1/sqrt3#

You can write the series as:

#sum_(n=0)^oo 3^nx^(2n) = sum_(n=0)^oo (3x^2)^n#

This is then a geometric series of ratio #3x^2#, which converges for:

#3x^2 < 1#

that is for:

#absx < 1/sqrt3#

and in such case the sum is:

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