Question

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

Answer 1

The radius of convergence is `R = e`

Explanation:

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`