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An example of a series involving factorials

  1. Apr 3, 2010 #1
    The problem at hand: [tex]\inline{\sum_{k=1}^n \frac{(k+1)!}{(k+3)!}}[/tex]
    Hence, find the limiting sum of the series, as n ---> infinity.

    Start this summation by expanding out the factorial to have a common factor of k!(k+1) as follows:

    [tex]\sum_{k=1}^n \frac{(k+1)!}{(k+3)!} = \sum_{k=1}^n \frac{k!(k+1)}{k!(k+1)(k+2)(k+3)} [/tex]

    Next step is to cancel the terms on the numerator and denominator:

    [tex]\sum_{k=1}^n \frac{k!(k+1)}{k!(k+1)(k+2)(k+3)} = \sum_{k=1}^n \frac{1}{(k+2)(k+3)}[/tex]

    Now consider the kth term:

    [tex]U_{k} = \frac{1}{(k+2)(k+3)}[/tex]

    Apply a differences method:

    [tex]U_{k} = \frac{1}{(k+2)(k+3)}*[\frac{(k+3) - (k+2)}{1}][/tex]

    [tex]U_{k} = \frac{(k+3)}{(k+2)(k+3)} - \frac{(k+2)}{(k+2)(k+3)}[/tex]

    [tex]U_{k} = \frac{1}{k+2} - \frac{1}{k+3}[/tex]

    [tex].: U_{k} = V_{k} - V_{k+1}[/tex]

    Note that [tex]\inline{V_{k} = \frac{1}{k+2} \forall{k} \in N}[/tex].

    [tex]S_{n} = U_{1} + U_{2} + U_{3} + U_{4} + ... + U_{n}[/tex]

    [tex]S_{n} = (V_{1} - V_{2}) + (V_{2} - V_{3}) + ... + (V_{n} - V_{n+1})[/tex]

    [tex].: S_{n} = V_{1} - V_{n+1}[/tex]

    [tex]\sum_{k=1}^n \frac{(k+1)!}{(k+3)!} = S_{n}[/tex]

    Now, substitute in the values k = 1 and k = (n + 1) into [tex]\inline{V_{k}}[/tex] to get the difference [tex]\inline{V_{1} - V_{n+1}}[/tex].

    [tex]V_{1} - V_{n+1} = \frac{1}{3} - \frac{1}{n+3}[/tex]

    [tex].: \sum_{k=1}^n \frac{(k+1)!}{(k+3)!} = \frac{1}{3} - \frac{1}{n+3}[/tex]

    It has therefore been shown that the answer to this sum to n terms is [tex]\inline{\frac{1}{3} - \frac{1}{n+3} \forall{n} \in N}[/tex], the result which can be proven by mathematical induction.

    And for the sum to infinity, follow these steps:

    [tex]\\lim_{n\Rightarrow {\infty}} \sum_{k=1}^n \frac{(k+1)!}{(k+3)!} = \\lim_{n\Rightarrow {\infty}} (\frac{1}{3} - \frac{1}{n+3})[/tex]

    [tex]\\lim_{n\Rightarrow {\infty}} (\frac{1}{3} - \frac{1}{n+3}) = \frac{1}{3}[/tex]

    [tex].: \\lim_{n\Rightarrow {\infty}} S_{n} = \frac{1}{3}[/tex]

    The problem has thus been completed.
  2. jcsd
  3. Apr 6, 2010 #2

    Gib Z

    User Avatar
    Homework Helper

    Is this for our general interest or something?
  4. Apr 6, 2010 #3
    Oh, sorry, wrote this out for someone, ended up posting it too, and I didn't explain its purpose.

    Could be taken as being for general interest if you like. :wink:
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