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Does this converge?

  1. Apr 10, 2007 #1
    1. The problem statement, all variables and given/known data

    I believe this problem diverges.

    n_infinity ne^-n^2

    heres the same equation: http://img100.imageshack.us/img100/6122/untitledig5.png

    2. Relevant equations



    3. The attempt at a solution

    upon pluggin in the n's as a sequence of numbers. it looks as it continues on without bound, so it would diverge, correct?

    ===========================

    also, for another problem...

    does this converge or diverge: 1 +1/8 + 2/27 + 2/64 + 1/125 + ...

    I first want to say that diverges because it continues on to infinity without bound.
    however, could i say instead that it converges to 0?

    i need an explanation for this one. Thankyou for any here.
     
    Last edited: Apr 10, 2007
  2. jcsd
  3. Apr 10, 2007 #2
    You mean as n goes to infinity
    [tex]ne^{-n^2}[/tex] goes to infinity?
    What happens to [tex]e^{-n}[/tex] as n goes to infinity?
     
    Last edited: Apr 10, 2007
  4. Apr 10, 2007 #3
    I see now that it's a series. Use the integral test to show it Converges.
    CC

    EDIT:Your second question:
    1+1/8+2/27+2/64+1/125 +.....can't go to zero because 1+(a bunch of positive things) can't be zero.
     
    Last edited: Apr 10, 2007
  5. Apr 10, 2007 #4
    is it safe to say the set of numbers, is divergent since it appears to continue on to infinity without bound?
     
  6. Apr 11, 2007 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, saying it appears to do anything is not sufficient (nor "safe"). I agree with happyg1: use the integral test. This series will converge as long as the corresponding integral
    [tex]\int_0^\infty xe^{-x^2}dx[/tex]
    converges and that is an easy integral.

    Notice that happyg1 gave the obvious answer: 1+ positive numbers can't converge to 0 because its partial sums are always larger than 1! You may be confusing this with the limit of the underlying sequence. If the sequence does not converge to 0, the sum cannot converge but here that sequence does converge to 0 which doesn't tell you anything.

    Do you recognize that your sum is
    [tex]\sum_1^\infty \frac{1}{n^3}[/itex]

    Do you know the "p-series" criterion? For what values of p does
    [tex]\sum_1^\infty \frac{1}{n^p}[/itex]
    converge?

    The integral test would also work nicely for this. Does
    [tex]\int_1^\infty \frac{1}{x^3} dx[/tex]
    converge?
     
  7. Apr 18, 2007 #6
    thankyou for the explanation.
     
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