How do you find the radius of convergence #Sigma 1/(n!)x^(n^2)# from #n=[1,oo)#?

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Answer 1 · 2017-02-04T00:02:55

The series:

#sum_(n=1)^oo x^(n^2)/(n!)#

has radius of convergence #R=1#

We can us the ratio test:

#abs (a_(n+1)/a_n) = abs (((x^((n+1)^2))/((n+1)!))/((x^(n^2))/(n!))) = abs(x)^(n^2+2n+1)/abs(x)^(n^2) (n!)/((n+1)!) = abs(x)^(2n+1)/(n+1)#

Clearly we have that:

#L= lim_(n->oo) abs (a_(n+1)/a_n) =lim_(n->oo) abs(x)^(2n+1)/(n+1) = {(0" for " abs(x) <=1),(oo" for " abs(x) >1):}#

So the radius of convergence is #R=1#