How do you find the radius of convergence $Sigma x^(3n)/5^n$ from $n=[1,oo)$ ?

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Answer 1 · 2017-02-16T09:46:52

The series:

$sum_(n=1)^oo x^(3n)/5^n$

is convergent for $x in [-root(3)5, root(3)5)$ , and absolutely convergent in the interior of the interval.

Write the series as:

$sum_(n=1)^oo x^(3n)/5^n = sum_(n=1)^oo (x^3/5)^n$

So, this is a geometric series of ratio: $x^3/5$ , which is convergent for:

$-1 <= x^3/5 < 1$

$-5 <= x^3 < 5$

$-root(3)5 <= x < root(3)5$

How do you find the radius of convergence Sigma x^(3n)/5^nSigma x^(3n)/5^n from n=[1,oo)n=[1,oo)?