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Homework Help: Factorial Sequences

  1. Aug 25, 2013 #1
    1. The problem statement, all variables and given/known data

    These trickly little buggers always seem to confuse me. I need to find out whether or not the sequence is increasing, decreasing or neither.


    2. Relevant equations

    3. The attempt at a solution

    I'm pretty sure that it's a decreasing sequences but when I expand and reduce, it's not so obvious:

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

    Which leaves me with,

    = (n!)/2n since one of the factorials in the numerator cancels out with the factorial in the denominator. Did I expand this wrong or something?

  2. jcsd
  3. Aug 25, 2013 #2
    Yes you expanded it wrong. (2n)! ≠ 2n*(n!).

    Try looking at An+1.
  4. Aug 25, 2013 #3
    Oops, double posted.
  5. Aug 25, 2013 #4
    Hey Mr. Anchovy,

    I stumbled upon what it expands to,

    (2n)! = 2ⁿ*(n!)*[ 1.3.5...(2n-1) ]

    And see hows it's decreasing.

    But how (2n)! expands out to that, seems unintuitive at this point.

    Using An+1 = ((n+1)!)^2/(2n+2)! Which still leaves me lost. I feel like I just don't understand factorials that well...
  6. Aug 25, 2013 #5


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    2ⁿ*(n!) = 2.4.6....(2n-2).2n
  7. Aug 26, 2013 #6
    Write out the first few and the last few terms of each series and match them up. But that is not going to help answer the question.

    Well (n+1)! = (n+1)n! and (2n+2)! = (2n+2)(2n+1)(2n)!, so write that expression with these substitutions and see if you can group terms together to get An+1 = (blob)An. If (blob) is less than 1 then the series is decreasing.
  8. Aug 26, 2013 #7


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    I love that, I will start using (blob) instead of c or n or x whenever I need a mathematical symbol!
    I'll probably even take out my old-fashioned ink pen just to be able to write (blob) by shaking it up and down.
  9. Aug 26, 2013 #8
    I wouldn't go that far, but in this case the more conventionally appropriate ## f(n) ## can easily cause confusion IME.
  10. Aug 26, 2013 #9
    Okay. I think I'm starting to see something, but am unsure if what I did was correct.

    ((n+1)n!)2/(2n+2)(2n+1)(2n) = (n!)2/(2n!)

    ((n+1)n!)2/(n!)2 = (2n+2)(2n+1)(2n!)/(2n!)

    (n+1)2 = (2n+2)(2n+1)

    n2+2n+1/4n2+5n+2 = 1

    I see that the denominator is greater from here and this will be less than 1. Did I do this right? Sorry if it's hard to read.
  11. Aug 26, 2013 #10
    Not really I'm afraid. It should go like this...[tex]
    A_n = \frac{n!^2}{(2n)!} \\
    A_{n+1} = \frac{(n+1)!^2}{(2(n+1))!} \\
    = \frac{((n+1)n!)^2}{(2n+2)(2n+1)(2n)!} \\
    = \frac{(n+1)^2 n!^2}{(2n+2)(2n+1)(2n)!} \\
    = \frac{(n+1)^2}{(2n+2)(2n+1)} . \frac{n!^2}{(2n)!} \\
    A_{n+1} = \frac{(n+1)^2}{(2n+2)(2n+1)}A_n[/tex]

    Now you can multiply out the brackets, and compare the numerator and denominator.

    I have taken the time to set this out because this is in general how you need to approach this kind of problem - write down an expression for ## A_{n+1} ##, expand it a bit, regroup terms to isolate ## A_n ## and see what you have got.
  12. Aug 26, 2013 #11
    Okay, that's some great knowledge that I can take forward. I'll just keep practicing on different problems and see what I get using what you showed me. Thanks!
  13. Aug 26, 2013 #12
    No problem. I can't stress enough how important this technique is - not just in its own right, but it forms the basis both of proof by induction and of analysis (the study of limits) which together probably make up about a quarter of the maths you will ever do!
  14. Aug 26, 2013 #13
    I'll be studying engineering and am thinking I won't do too much in terms of analysis, unless of course I want to.

    Another method in my text, which I somehow missed, was to take the ratio of A_n/A_(n+1) then simplify and take the limit as n approaches infinite. If the limit is infinite then it's eventually decreasing. Which seems like it's along the lines of what you had me do. Both have worked great so far!
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