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Infinite Series using Falling Factorials

  1. Sep 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine [itex]\sum_{k=0}^\infty \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}[/itex].


    2. Relevant equations
    ## (x)_m=x(x-1)(x-2)...(x-(m-1)) ##, integer ##m\geq0##.
    ## (x)_{-m}=\frac{1}{(x+1)(x+2)...(x+m)}##, integer ##m>0##.
    ## Δ((x)_m)=m(x)_{m-1}##
    [itex]\sum_{a\leq k<b} (k)_m=\frac{(x)_{m+1}}{m+1} |_a^b[/itex], if ##m\neq-1##.

    3. The attempt at a solution
    First I noted that [itex](4k)_{-4}=\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)} [/itex]; so I rewrote the series as $$ \lim_{n\rightarrow \infty}(\sum_{k=0}^n (4k)_{-4}). $$
    Evaluating the series I get,
    [tex]
    \begin{align}
    \sum_{k=0}^n (4k)_{-4} &=\frac{(4k)_{-3}}{-3}|_0^{n+1}\\
    &= \frac{(4(n+1))_{-3}}{-3}-\frac{(0)_{-3}}{-3}\\
    &= -\frac{1}{3}\frac{1}{[4(n+1)+1][4(n+1)+2][4(n+1)+3]}+\frac{1}{18}
    \end{align}
    [/tex]
    I then took the limit of the closed form expression for the series above, $$ \lim_{n\rightarrow \infty}(-\frac{1}{3}\frac{1}{[4(n+1)+1][4(n+1)+2][4(n+1)+3]}+\frac{1}{18})=\frac{1}{18}.$$

    The answer is ln2/4-pi/24. This is the first time that I've used the falling factorials outside of the textbook in which they appeared; so I'm not sure if using it on this question is even correct (but i can't see why it wouldn't be). If anyone can give me a hint or explanation, it would be much appreciated.
     
  2. jcsd
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