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Another 'Does this series converge?'

  1. Feb 10, 2008 #1
    1. The problem statement, all variables and given/known data

    The sum from n=1 to infinity of n!*2^n*n^(-n)

    2. Relevant equations



    3. The attempt at a solution

    not a clue.
     
  2. jcsd
  3. Feb 10, 2008 #2
    try using the ratio test
     
  4. Feb 10, 2008 #3
    you should first make the series simple:

    The series you have is the same as: the sum from n=1 to infiniti of
    n!*(2/n)^n "because n^(-n) is the same as 1/n^(n)).

    Now you can use the ratio test:
    take the limite as n approaches infiniti of [(n+1)!*(2/n)^(n+1)] / [n!*(2/n)^n ]
    this limit becomes after simplifying:
    limit as n approaches infiniti of (2n+2)/n
    you can fact the n out and cancel it out with the n in the denominator to get:
    limit as n approaches infiniti of 2+(2/n) which equals 2
    so because 2 is grater than one, this series converges by the ratio test
     
  5. Feb 10, 2008 #4
    No, when you simplify that ratio, you get 2(n/(n+1))^n, whose limit is 2/e. This is less than 1, so the series CONVERGES.

    If you get 2 as the limit, the ratio test says that it DIVERGES.
     
  6. Feb 10, 2008 #5
    (2n+2)/n
    =n*(2+2n^-1)/n
    so now you cancel out the n in the denominator with the one above it
    you get:
    2+(2/n)

    when you take the limit as n approaches infiniti you get 2
     
  7. Feb 12, 2008 #6
    how do you get this?

    i know that (1 + c/n)^n -> e^c as n-> infin
     
  8. Feb 12, 2008 #7
    but how do you get 2/e, well c= -1, but how do you get that?
     
  9. Feb 12, 2008 #8
    Alright,

    the series you have is n!*2^n*n^-n
    this can be simplifyed to become:

    An= n!*(2/n)^n

    so An+1= (n+1)!*(2/n)^(n+1) "you just replace n by n+1

    The ratio test says lim as n goes to infinity of An+1/An
    which is the same as:

    lim as n aprroaches infiniti of [(n+1)!*(2/n)^(n+1)] / [n!*(2/n)^n]

    (n+1)!/n! equals n+1 and (2/n)^(n+1)/(2/n)^(n) equals 2/n

    so the limit becomes:

    lim as n approaches infiniti of (n+1)*(2/n)

    which is the same as:

    lim as n approaches infiniti of (2n+2)/n
    you can factor out the

    The limit becomes:

    n*(2+2*n^-1)/n

    cancel out the n in the denominator and the one above it:

    now you have the limit as n aprroaches infiniti of 2+(2/n) or 2+2*n^-1

    which equals 2
     
  10. Feb 12, 2008 #9

    Dick

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    Wrong again. Not, "An+1= (n+1)!*(2/n)^(n+1)". An+1=(n+1)!*(2/(n+1))^(n+1). Mattofix, can you correct the rest? As masnevets has already pointed out, the ratio test limit is 2/e.
     
  11. Feb 12, 2008 #10
    yeah - i know that torresmido is wrong.

    i have handed it in already though.

    but i got to 2(n/(n+1))^n and then didnt know how to get 2/e as the limit.

    thanks for your help guys
     
  12. Feb 12, 2008 #11

    Dick

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    Write it as 2/((n+1)/n)^n=2/(1+1/n)^n. Look familiar now?
     
  13. Feb 12, 2008 #12
    Sorry guys ... I see my mistake now.
    the limimit is 2/e so the series diverges by the ratio test
     
  14. Feb 12, 2008 #13

    Dick

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    You mean converges, right? 2/e<1.
     
  15. Feb 12, 2008 #14
    oh - yeah, now it makes sense
     
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