Question

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

Answer 1

The interval of convergence is `x=0` and the radius of convergence is `R=0`.

Explanation:

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

The series `suma_n` is convergent when `L<1` when `L=lim_(nrarroo)abs(a_(n+1)/a_n)`. Here, this becomes:

`L=lim_(nrarroo)abs(((n+1)!(x^(n+1)))/(n+1)^(1/2(n+1))*n^(n/2)/(n!(x^n)))`

Simplifying:

`L=lim_(nrarroo)abs(((n+1)(n!))/(n!)*x^(n+1)/x^n*n^(n/2)/(n+1)^(n/2+1/2))`

`L=lim_(nrarroo)abs((n+1)/(n+1)^(n/2+1/2)*x*n^(n/2))`

The `x` can be moved to outside the limit, since the limit only depends on `n`.

Also note that `(n+1)/(n+1)^(n/2+1/2)=(n+1)^(1-(n/2+1/2))=(n+1)^(1/2-n/2)`.

`L=absxlim_(nrarroo)abs((n+1)^(1/2-n/2)(n^(n/2)))`

`L=absx(lim_(nrarroo)abs((n+1)(n/(n+1))^n))^(1/2)`

Note that `lim_(nrarroo)((n+1)/n)=e`, so `lim_(nrarroo)(n/(n+1))=1/e`. Thus, the overall limit approaches `oo` since the remaining portion of `n+1` is unbounded.

Since the limit approaches infinity, the only time when `L<1` will be when `x=0`, as this makes `L=0`.

Thus the interval of convergence is `x=0` and the radius of convergence is `R=0`.