Calculus  ratio test
posted by COFFEE .
infinity of the summation n=1: (e^n)/(n!) [using the ratio test]
my work so far:
= lim (n>infinity)  [(e^n+1)/((n+1)!)] / [(e^n)/(n!)] 
= lim (n>infinity)  [(e^n+1)/((n+1)!)] * [(n!)/(e^n)] 
= lim (n>infinity)  ((e^n)(e^1)(n!)) / ((n+1)(n!)(e^n)) 
..the e^n & n! cancels out
= lim (n>infinity)  (e^1) / (n+1) 
im stuck here.. how do i finish this?
and also to find out if it's Divergent (L>1), convergent (L<1), or fails (L=1)?
The limit is zero. So, L = 0 therefore the series is convergent.
The value of the summation is e^e  1
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