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Practice - Series

  1. Nov 3, 2007 #1
    Ok, first

    I am face with the series from 1 to infinity of (n!)e^(-n)

    I used the ratio test and came out with lim-->infinity of (n+1)e through cancellation.

    I don't see how the series is divergent, can anyone explain?



    Secondly, for the series from 1 to infinity of 11/n(n+2) -- > I can determine that it is convergent using the limit test with 1/n to show that the limit is a number > 0 so both diverge.

    But how can I get the sum??????

    Someone pointed out that this is a telescoping series. But how can I know this? and how do I get the partial fractions.


    This will help me prepare for my coming exam. THANK YOU.
     
  2. jcsd
  3. Nov 3, 2007 #2
    in order for a series to diverge, the result of the ratio test must be greater than 1
    since im-->infinity of (n+1)e
    is really just infinity * e (e is a constant), the series = infinity
    therefore, the series is divergent
     
  4. Nov 3, 2007 #3
    I see, i thought the limit had to be a number.

    could anyone help with 2?
     
  5. Nov 4, 2007 #4

    HallsofIvy

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    No, you don't. Since the limit is as n goes to infinity, the limit cannot depend on n. The limit is actually infinity (i.e. the sequence does not converge itself). Since that is certainly NOT less than 1, the series diverges.

    You don't? One of the first things you should have learned about series is that if the sequence {an} does not converge to 0, the series cannot converge. Here, in fact, (n!)e-n goes to infinity itself! This series is very badly divergent.


    I have no idea what you are talking about here! It is fairly easy to see that 11/n(n+2)< 11/n2 and the latter converges because the power of n is greater than 1.

    But when you say " can determine that it is convergent using the limit test with 1/n to show that the limit is a number > 0 so both diverge" you have lost me! You can't show that a sequence converges by comparing it to a divergent series. And what do you mean when you say "so both diverge"? You had already come to the conclusion that the sequence converges!

    If you know what a "telescoping series" is, that should be easy. Write out a few terms (I am going to drop the 11 since you can always multiply by 11 afterward) 1/n(n+2) gives, for n= 1, 1/1(3)= 1/3= 1/2- 1/6, for n= 2, 1/2(4)= 1/8= 1/4- 1/8, for n= 3, 1/3(5)= 1/15= 1/6- 1/10, 1/4(6)= 1/8- 1/12, 1/5(7)= 1/10- 14 etc. You should be able to show that 1/n(n+2)= 1/(2n)- 1/(2(n+2)).
    Now put them into the series: (1/2- 1/6)+ (1/4-1/8)+ (1/6 - 1/10)+ (1/8- 1/12)+ (1/10- 1/14)+ ... Do you see that the first "-1/6" is canceled by a later "+1/6", that "-1/8" is canceled by a later "+1/8", that "-1/10" is canceled by a later "+1/10" and that, in fact, each "-1/(2n)" is canceled by a later "+1/(2(n+2))"? That's what is meant by "telescoping series"- all terms cancel so that the series "closes up" like a collapsible telescope. For any finite partial sum, to N, the only parts that don't cancel are the first 1/2 and the last -1/(2N): the Nth partial sum is 1/2- 1/(2N). What is the limit of that as N goes to infinity?

    ?? Yes, a limit (of a numerical sequence or series) has to be a number. But that is only if it HAS a limit (if it converges).
     
    Last edited: Nov 4, 2007
  6. Nov 4, 2007 #5
    I guess the thing that is throwing me off here is the partial fraction part, here is what I am getting:

    11 = A(n+2) + Bn

    Then, setting n = -2, I get:

    11 = 0 + -2B
    and B = 11/-2

    --> can anyone explain what I am doing wrong? thank you in advance.
     
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