How do you show the convergence of the series #(n!)/(n^n)# from n=1 to infinity??

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Answer 1 · 2018-05-16T00:34:31

The series converges

The #n#-th term of the series is #t_n = (n!)/n^n#. Hence we have

#t_(n+1)/t_n = ((n+1)!)/(n+1)^(n+1) times n^n/(n!)#
#qquad = (n+1)/(n+1)^(n+1) n^n = (n/(n+1))^n#
#qquad = 1/(1+1/n)^n#

Now, it is well known that #lim_{n to oo} (1+1/n)^n = e# (indeed, that's the definition of the number #e#). And thus

#lim_{n to oo }|t_{n+1}/t_n| = 1/e <1#

and thus the series converges according to the ratio test.