Question #822e7

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Answer 1 · 2018-01-26T10:05:55

The series converges.

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

Since #e/3<1#, the series converges.