How do you apply the ratio test to determine if #Sigma (3^n(n!))/(n^n)# from #n=[1,oo)# is convergent to divergent?

We can apply d'Alembert's ratio test:

Suppose that;

# S=sum_(r=1)^oo a_n \ \ #, and #\ \ L=lim_(n rarr oo) |a_(n+1)/a_n| #

Then

if L < 1 then the series converges absolutely;
if L > 1 then the series is divergent;
if L = 1 or the limit fails to exist the test is inconclusive.

So our series is;

# S=sum_(n=1)^oo (3^n(n!))/n^n #

So our test limit is:

# L = lim_(n rarr oo) | {(3^(n+1)((n+1)!))/(n+1)^(n+1)}/{(3^n(n!))/n^n}| #
# \ \ \ = lim_(n rarr oo) | (3^(n+1)(n+1)!)/(n+1)^(n+1) * n^n/(3^n n!}| #
# \ \ \ = lim_(n rarr oo) | (3 * 3^n(n+1) * n!)/(n+1)^(n+1) * n^n/(3^n n!}| #
# \ \ \ = lim_(n rarr oo) | (3 (n+1))/(n+1)^(n+1) * n^n| #
# \ \ \ = lim_(n rarr oo) | (3n^n )/(n+1)^n | #
# \ \ \ = 3 \ lim_(n rarr oo) | (n/(n+1))^n | #
# \ \ \ = 3 \ lim_(n rarr oo) (n/(n+1))^n #
# \ \ \ = 3/e #
# \ \ \ ~~ 1.1 > 1 #

Hence the series is divergent.

In case you are wondering how the final limit was established; it is a standard limit in disguise:

Let #u=n+1 => n=u-1# then #u rarr oo# as #n rarr oo#, And

# lim_(n rarr oo) (n/(n+1))^n = lim_(u rarr oo) ((u-1)/u)^(u-1)#
# " " = lim_(u rarr oo) (1-1/u)^u / (1-1/u)#
# " " = {lim_(u rarr oo) (1-1/u)^u }/ {lim_(u rarr oo)(1-1/u)}#
# " " = lim_(u rarr oo) (1-1/u)^u #

Which courtesy of Leonhard Euler is standard limit with known value #1/e#.

Using Strirling approximation

#log n! =nlogn-n+O(logn)#

we have

#(3^n(n!))/(n^n) approx (3/e)^n# but as #3/e > 1# the series diverges.