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 $R=1$

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