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

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Answer 1 · 2017-02-13T16:14:58

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

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#.