# Infinite Series using Falling Factorials

1. Sep 28, 2012

### WittyName

1. The problem statement, all variables and given/known data
Determine $\sum_{k=0}^\infty \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}$.

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 $m>0$.
$Δ((x)_m)=m(x)_{m-1}$
$\sum_{a\leq k<b} (k)_m=\frac{(x)_{m+1}}{m+1} |_a^b$, if $m\neq-1$.

3. The attempt at a solution
First I noted that $(4k)_{-4}=\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}$; so I rewrote the series as $$\lim_{n\rightarrow \infty}(\sum_{k=0}^n (4k)_{-4}).$$
Evaluating the series I get,
\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}
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.