Trying to find #sum_(n=1)^oo(n^n/(n!*3^n))# (I'm staring at #n=1# instead of #n=0# because of the #0^0# when #n=0#.)
Using the Ratio Test:
#lim_(n to oo)|(n+1)^(n+1)/((n+1)!\*3^(n+1))*(n!\*3^n)/n^n|#
#=lim_(n to oo)|((n+1)(n+1)^(n))/((n+1)\*n!\*3^(n)\*3)*(n!\*3^n)/n^n|#
#=lim_(n to oo)|(n+1)^(n)/(3n^n)|#
#=1/3lim_(n to oo)|((n+1)/n)^(n)|#
#=1/3lim_(n to oo)(1+1/n)^(n)#
#=1/3*e# because #lim_(n to oo)(1+1/n)^(n)# is famously equal to #e#. See this video for the work.
Since #e/3<1#, the series converges.