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:

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

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