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#