Question

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

Answer 1

The radius of convergence is `R=1/27`.

Explanation:

We can use the ratio test. Evaluate the ratio:

`abs(a_(n+1)/a_n) = abs ( (((3(n+1))!)/(((n+1)!)^3) x^(n+1)) / (((3n)!)/(n!)^3x^n)`

`abs(a_(n+1)/a_n) = (((3(n+1))!)/((3n)!) (n!)^3/ ((n+1)!)^3)abs(x)`

`abs(a_(n+1)/a_n) = ((3n+3)!)/((3n)!) 1/ (n+1)^3abs(x)`

`abs(a_(n+1)/a_n) = ( (3n+3)(3n+2)(3n+1) )/(n+1)^3abs(x)`

So we have:

`lim_(n->oo) abs(a_(n+1)/a_n) = 27abs(x)`

Which means that the series is absolutely convergent for `abs(x)< 1/27` and that the radius of convergence is `R=1/27`.