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

1 Answer
Mar 2, 2017

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#.