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Converge or diverge of factorials

  1. Mar 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine whether the series converges or diverges.

    [tex]\sum\frac{37}{n!}[/tex]


    I will just forget about the 37, and think of it as [tex]\sum\frac{1}{n!}[/tex]

    I can try to decompose the n!

    n! = n(n-1)!
    n! = n ( n-1) (n-2)!
    n! = n(n-1)(n-2)(n-3)......2*1


    so [tex]\sum\frac{1}{n!}[/tex] = [tex]\sum\frac{1}{n(n-1)!}[/tex]

    =

    [tex]\frac{1}{n}[/tex] *[tex]\frac{1}{(n-1)!}[/tex]

    since 1/n is a p-series and diverges so does the series.

    Is this right ?
     
  2. jcsd
  3. Mar 22, 2009 #2

    Dick

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    That's not a p-series. A p-series looks like 1/n^p. Use a ratio test.
     
    Last edited: Mar 22, 2009
  4. Mar 22, 2009 #3
    Don't know what a ratio test is. Isn't 1/n a p-series because

    1/n = 1/n^1 ?
     
  5. Mar 22, 2009 #4
    Your series isnt 1/n but 1/n!, can you see the difference?
     
  6. Mar 22, 2009 #5
    yes but

    its 1/n * 1 / (n-1)!

    if one term diverges, does not the other because Infinity * whatever != converge
     
  7. Mar 22, 2009 #6
    Well, nope!

    1/n*1/n^2=1/n^3. this series obviously converges, by applying p-test

    Eventhough the first one alone(the harmonic series diverges)

    Moreover: 1/n*1/n both alone diverge but when multiplied together they converge , 1/n^2 according to p-test again.
     
  8. Mar 22, 2009 #7
    ok, so how do I go about this. Since I haven't learned ratio test yet
     
  9. Mar 23, 2009 #8
    I assume you have learned the comparison tests, right? since this is one of the first tests you learn when you are introduced to numerical series.

    first prove that:

    [tex]n!>2^n[/tex] for some n>k, where k is a positive integer, and then use this fact and the comparison theorme to show that your series converges.
     
  10. Mar 23, 2009 #9
    From using a graphical interface k = 4.

    not sure how I can prove it.

    I could try this :(induction)

    0! = 1
    1! = 1
    2! = 2
    3! = 6
    4! = 24
    5! = 120

    2^1 = 2
    2^2 = 4
    2^3 = 8
    2^4 = 16
    2^5 = 32
     
  11. Mar 23, 2009 #10
    well, since n!>2^n, for say n>2, then it follows that

    1/n!<(1/2)^n, and you probbably know that the RHS is a geometric sequence whose ratio is r=1/2<1, so it converges, now from the comparison test, we know that the series sum(1/n!) converges as well .
     
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