Saturday

October 25, 2014

October 25, 2014

Posted by **COFFEE** on Sunday, July 29, 2007 at 11:24pm.

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) |

**Answer this Question**

**Related Questions**

calculus - ratio test - infinity of the summation n=1: (e^n)/(n!) [using the ...

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 - please help me...for the power series summation of n^4x^n/2(n-1)factorial...

Calculus 2 - In the following series x is a real number. In each case, use the ...

CALC 2 - In the following series x is a real number. In each case, use the ratio...

CALC 2 pls help!! - In the following series x is a real number. In each case, ...

Calculus - How do I find the radius of convergence of a series where n=1 to ...

CALC 2 - a. Consider the following limit as a fact: lim n-> infinity ((n!)^1/...