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)?
For Further Reading
* Calculus  ratio test  Count Iblis, Sunday, July 29, 2007 at 7:01pm
The limit is zero. So, L = 0 therefore the series is convergent.
The value of the summation is e^e  1

thank you for your response, but..
how did you get the summation of e^(e1)
...from this?
= lim (n>infinity)  (e^1) / (n+1) 
i think i figured it out.
Respond to this Question
Similar Questions

Calculus  ratio test
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!)) … 
calculus  ratio test
Posted by COFFEE on Sunday, July 29, 2007 at 6:32pm. 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)!)] … 
calculus  interval of convergence
infinity of the summation n=0: ((n+2)/(10^n))*((x5)^n) .. my work so far. i used the ratio test = lim (n>infinity)  [((n+3)/(10^(n+1)))*((x5)^(n+1))] / [((n+2)/(10^n))*((x5)^n)]  .. now my question is: was it ok for me to … 
calculus  interval of convergence
infinity of the summation n=0: ((n+2)/(10^n))*((x5)^n) .. my work so far. i used the ratio test = lim (n>infinity)  [((n+3)/(10^(n+1)))*((x5)^(n+1))] / [((n+2)/(10^n))*((x5)^n)]  .. now my question is: was it ok for me to … 
Calc. Limits
Are these correct? lim x>0 (x)/(sqrt(x^2+4)  2) I get 4/0= +/ infinity so lim x>0+ = + infinity? 
calc
Are these correct? lim x>0 (x)/(sqrt(x^2+4)  2) I get 4/0= +/ infinity so lim x>0+ = + infinity? 
Calc Please Help
Are these correct? lim x>0 (x)/(sqrt(x^2+4)  2) I get 4/0= +/ infinity so lim x>0+ = + infinity? 
Calculus
Find the horizontal asymptote of f(x)=e^x  x lim x>infinity (e^x)x= infinity when it's going towards infinity, shouldn't it equal to negative infinity, since 0infinity =  infinity lim x> infinity (e^x)x= infinity 
Math
1. If 1/infinity = infinity or infinity ? 
Math
1. If 1/infinity = infinity or infinity ?