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Factorials & Series - im a little lost here

  1. Aug 9, 2006 #1
    I Have a small problem.

    I know the following:

    The series:
    S_{n} = 1 + 2 + 3 + 4 + 5 = \sum_{i=1}^{n=5}i

    2S_{n} = 1 + 2 + 3 + 4 + 5 + 1 + 2 + 3 + 4 + 5 = n(n+1) = n^{2} + n

    S_{n} = \frac{2S_{n}}{2} = \frac{n^{2} + n}{2}

    \sum_{i=1}^{n=5}i = \frac{n^{2} + n}{2} = \frac{25 + 5}{2} = 15

    Ok, thats simple.
    But i was attempting to do the same with the factorials, where:

    ln(n!) = ln(5!) = ln(1) + ln(2) + ... + ln(5) = \sum_{i=1}^{n=5}log(i)

    Is it possible to apply the same method, or something similar here? I did try a few but got a bit cheesed off.

    In the first case, you just duplicate and reverse the sequence and add to get a constant sequence.
    Last edited: Aug 9, 2006
  2. jcsd
  3. Aug 9, 2006 #2
    [tex]ln(n!) = ln(5!) = ln(1) + ln(2) + ... + ln(5) = \sum_{i=1}^{n=5}ln(i)[/tex]

    Don't you mean that? Sum of Ln(i) ?

    [tex]2\sum_{i=1}^{n=5}i[/tex] can also be written as [tex]\sum_{i=1}^{n=5}2i[/tex] if I'm not mistaking. Your algebration wrote it as [tex]\sum_{i=1}^{n=5}i + \sum_{i=1}^{n=5}i[/tex] which is correct but not necessarily useful.

    Now...If I recall the Logaritmic properties, Log(a)+Log(b)=Log(ab)

    So [tex]ln(n!)= Ln(1*2*3*4*5...n) = Ln(1)+Ln(2)+Ln(3)...+Ln(n)=\sum_{i=1}^{n=oo}ln(i)[/tex]

    From what I can tell you are correct. Well except n=5 part. If you have
    [tex]ln(n!)[/tex] without any other condition it will equal [tex]\sum_{i=1}^{n=oo}ln(i)[/tex]
    Last edited: Aug 9, 2006
  4. Aug 9, 2006 #3
    Yes, thanks.

    Erm, yeah, working the other way though :P

    The idea is, one fast method for calculating the sum of the positive integers to a given value, say 10, is to take 0.5(10^2 + 10).

    Well i was wondering if there is a shortcut for factorials in a similar manner.

    Its late, and the coffee has stopped working....
  5. Aug 9, 2006 #4
    That is true...but all I got in my Pre-Calculus about sums is some basic ways to do them, no explanations.

    (n^2+n)/2 if you have a sum of an unknown with power of 1 starting at 1, (n^2-n)/2 if it starts at zero I think...and some formulas for situations when you make a Sum of...n^2 and n^3 etc...But never why.

    They showed some substraction between terms, double substraction if its quadratic until the difference is the same...something about a 3x4 matrix for quadratics...If I stretch my brain i may be able to reproduce it but it didn't explain most of anything. That is one area of Mathematics that I'd like to revisit in the future.
  6. Aug 9, 2006 #5


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    there is no exact formula, but there is a very good approximation for [itex] \log(n!) [/itex] that gets better as [itex] n [/itex] gets large. it's called Stirling's approximation and is:

    [tex] \frac{1}{12 n + 1} < \log \left( \frac{n!}{\sqrt{2 \pi n}} \right) - n \left( \log(n) - 1 \right)< \frac{1}{12 n} [/tex]


    [tex] n \left( \log(n) - 1 \right) + \frac{\log(2 \pi n)}{2} + \frac{1}{12 n + 1} < \log(n!) < n \left( \log(n) - 1 \right) + \frac{\log(2 \pi n)}{2} + \frac{1}{12 n} [/tex]

    i would suggest to Google or Wikipedia it to find out why this is true.
    Last edited: Aug 9, 2006
  7. Aug 10, 2006 #6


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    You might want to look up Euler-Maclaurin summation.
  8. Aug 10, 2006 #7
    Thanks, got a fresh day and a new cup of coffee. Ill read up on those two topics.
  9. Apr 6, 2010 #8
    You can get back to the nth factorial with this representation:

    [tex]e^{\sum_{k=1}^n ln(k)}[/tex]

    Not sure if this is at all helpful to you, but there's a statement that returns you to n!

    Try it out, I assure you it works, except if you let k= 0 as ln(0) is undefined.
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