How do you find the radius of convergence #Sigma x^(6n)/(n!)# from #n=[1,oo)#?

Question

2963 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-27T14:16:56

The answer is #-oo < x < +oo #

Apply the ratio test to compute the convergence interval

#lim_(n->oo)|((x^(6(n+1))/((n+1)!))/(x^(6n)/(n!)))|#

#=lim_(n->oo)(|x^6/(n+1)|)#

#=|x^6|lim_(n->oo)(|1/(n+1)|)#

#=|x^6|*0#

#=0#

As the limit is #<1# for every #x#, therefore #sum_(n=1)^(+oo)x^(6n)/(n!)# converges for all #x#