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

Question

1730 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-07T14:33:37

The radius of convergence is $R = e$

Using the ratio test:

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

$abs (a_(n+1)/a_n) = ((n+1)!)/(n!) xx n^n/(n+1)^(n+1) xx abs (x^(n+1)/x^n)$

$abs (a_(n+1)/a_n) = (n+1) xx 1/(n+1) (n/(n+1))^n xx abs (x)$

$abs (a_(n+1)/a_n) =abs (x)/((n+1)/n)^n$

$abs (a_(n+1)/a_n) =abs (x)/(1+1/n)^n$

So:

$lim_(n->oo) abs (a_(n+1)/a_n) = absx lim_(n->oo) 1/(1+1/n)^n = absx/e$

So the radius of convergence is $R = e$ which means that the series is absolutely convergent for $absx < e$

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