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!)) / ((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.

  1. 👍 0
  2. 👎 0
  3. 👁 129
asked by COFFEE

Respond to this Question

First Name

Your Response

Similar Questions

  1. 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) |

    asked by COFFEE on July 29, 2007
  2. 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) |

    asked by COFFEE on July 29, 2007
  3. CALC 2 pls help!!

    In the following series x is a real number. In each case, use the ratio test to determine the radius of convergence of the series. Analyze the behavior at the endpoints in order to determine the interval of convergence. a.

    asked by BAE on April 14, 2014
  4. CALC 2

    In the following series x is a real number. In each case, use the ratio test to determine the radius of convergence of the series. Analyze the behavior at the endpoints in order to determine the interval of convergence. a.

    asked by Bae on April 13, 2014
  5. Calculus 2

    In the following series x is a real number. In each case, use the ratio test to determine the radius of convergence of the series. Analyze the behavior at the endpoints in order to determine the interval of convergence. a.

    asked by BAE on April 14, 2014
  6. Calculus

    Which of the following series could be tested for convergence/divergence with the integral test? the summation from n=1 to infinity of 1/n! the summation from n=1 to infinity of 1/n the summation from n=2 to infinity of ln(n)/n^2

    asked by Alice on May 13, 2019
  7. calculus - interval of convergence

    infinity of the summation n=0: ((n+2)/(10^n))*((x-5)^n) .. my work so far. i used the ratio test = lim (n-->infinity) | [((n+3)/(10^(n+1)))*((x-5)^(n+1))] / [((n+2)/(10^n))*((x-5)^n)] | .. now my question is: was it ok for me to

    asked by COFFEE on July 30, 2007
  8. calculus - interval of convergence

    infinity of the summation n=0: ((n+2)/(10^n))*((x-5)^n) .. my work so far. i used the ratio test = lim (n-->infinity) | [((n+3)/(10^(n+1)))*((x-5)^(n+1))] / [((n+2)/(10^n))*((x-5)^n)] | .. now my question is: was it ok for me to

    asked by COFFEE on July 29, 2007
  9. CALC

    please help me...for the power series summation of n^4x^n/2(n-1)factorial....find the radius and interval of convergence... so far i am guessing that radius is equal to x and interval of convergence is (-infinty to positive

    asked by Maria on May 7, 2007
  10. Calc 2, Series/Sequences

    Explain clearly in detail the difference of the summation of n=1 to infinity of 1/n^2; and the integral of 1/x^2dx with the limit of 1 to infinity, as well. Part 2: Explain the natures of the two expressions and explain why the

    asked by Alex on April 1, 2010

More Similar Questions