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

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Answer 1 · 2017-01-20T23:06:13

The radius of convergence is #R=sqrt(e)#

We can apply the ratio test, stating that a series:

#sum_(n=1)^oo a_n#

is absolutely convergent if:

#lim_(n->oo) abs (a_(n+1)/a_n) < 1#

Calculate the expression of the ratio for this series:

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

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

Now we have:

#lim_(n->oo) (n/(n+1))^n = lim_(n->oo) 1/(((n+1)/n)^n) = lim_(n->oo) 1/((1+1/n)^n) = 1/e#

so that:

#lim_(n->oo) abs (a_(n+1)/a_n) = x^2/e#

The series is then absolutely convergent if:

#x^2/e < 1 => abs(x) < sqrt(e)#

which means the radius of convergence is #R=sqrt(e)#