Posted by **COFFEE** on Monday, July 30, 2007 at 6:16pm.

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^(e-1)

...from this?

= lim (n->infinity) | (e^1) / (n+1) |

i think i figured it out.

## Answer This Question

## Related Questions

- calculus - ratio test - Posted by COFFEE on Sunday, July 29, 2007 at 6:32pm. ...
- Calculus - ratio test - infinity of the summation n=1: (e^n)/(n!) [using the ...
- calculus - interval of convergence - infinity of the summation n=0: ((n+2)/(10^n...
- calculus - interval of convergence - infinity of the summation n=0: ((n+2)/(10^...
- Calc. Limits - Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0...
- calc - Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- ...
- Calc Please Help - Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/...
- Calculus - Find the horizontal asymptote of f(x)=e^x - x lim x->infinity (e^x...
- Math - 1. If -1/infinity = infinity or -infinity ? 2. If lim x->infinity^- = ...
- Math - 1. If -1/infinity = infinity or -infinity ? 2. If lim x->infinity^- = ...

More Related Questions