Does #a_n=n^n/(n!) # converge?
2 Answers
Nov 5, 2015
No, but
Explanation:
#a_(n+1) = (n+1)^(n+1)/((n+1)!)#
#=(((n+1)/n)^n n^n (n+1))/(n!(n+1))#
#=((1+1/n)^n n^n)/(n!) = (1+1/n)^n a_n#
Now
So
Nov 5, 2015
In addition, the fact that
Explanation:
The Ratio Test says that a series