Infinite Series using Falling Factorials I recently started reading Concrete Mathematics by Graham, Knuth and Patashnik and met falling/rising factorials for the first time; it seemed like a very convenient method for evaluating particular sums.
I saw the following problem asked on a different site and tried to use falling factorials to get the answer:$$\text{Determine}\ \sum_{k=0}^\infty \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}.$$
Here's what i did:
First I noted that $(4k)^\underline{-4}=\dfrac{1}{(4k+1)(4k+2)(4k+3)(4k+4)} $; so I rewrote the series as $$ \lim_{n\rightarrow \infty}\left(\sum_{k=0}^n (4k)^\underline{-4}\right)\;.$$
Evaluating the series I get,
\begin{align}
\sum_{k=0}^n (4k)^\underline{-4} &=\frac{(4k)^\underline{-3}}{-3}\Bigg|_0^{n+1}\\
&= \frac{[4(n+1)]^\underline{-3}}{-3}-\frac{0^\underline{-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, $$ \lim_{n\rightarrow \infty}\left(-\frac{1}{3}\frac{1}{[4(n+1)+1][4(n+1)+2][4(n+1)+3]}+\frac{1}{18}\right)=\frac{1}{18}\;.$$
$ \frac{1}{18} $ is apparently the wrong answer. This is the first time that I've used the falling factorials outside of the textbook, so I'm not sure if using it on this question is 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. 
 A: Completely revised and corrected:
The problem is that you need a chain rule to handle the factor of $4$ in $4k$, and the finite calculus doesn’t have one. In general it’s very hard to deal with composite functions other than simple shifts by adding a constant to the variable. To see the problem, consider what function $f(x)$ would have $\Delta f(x)=(2x)^{\underline 2}$. Analogy with ordinary calculus would suggest that it ought to be something like $(2x)^{\underline 3}$, probably adjusted by a constant coefficient. But 
$$\begin{align*}
\Delta(2x)^{\underline 3}&=\big(2(x+1)\big)^{\underline 3}-(2x)^{\underline 3}\\
&=(2x+2)(2x+1)(2x)-(2x)(2x-1)(2x-2)\\
&=4x\Big((x+1)(2x+1)-(2x-1)(x-1)\Big)\\
&=24x^2\;,\tag{1}
\end{align*}$$
while $(2x)^{\underline 2}=(2x)(2x-1)=4x^2-2x$; clearly these are not constant multiples of each other. Multiplying $(1)$ by $\frac16$, as is suggested by ordinary calculus, gets the $x^2$ term right, but it does nothing to fix give us the missing first power term. Now $\Delta x^{\underline 2}=2x$, so in this case we actually can fix things up:
$$\Delta\left(\frac16(2x)^{\underline 3}-x^{\underline 2}\right)=\frac16\left(24x^2\right)-2x=4x^2-2x=(2x)^{\underline 2}\;.$$
But that was an ad hoc fix-up after the fact that doesn’t obviously point the way to a general solution even in the case of a positive exponent.
I don’t know a slick finite calculus method for evaluating your limit, but this at least explains why what you tried didn’t work.
A: You might try writing , 1/((4k+1)(4k+2)(4k+3)(4k+4))
=1/(6(4k+1)) + 1/(2(4k+3)) - 1/(4(2k+1)) - 1/(24(k+1))
A: As Brian M. Scott already pointed out, you need a way to handle the factor 4, or more general, you need a rule for to sum/difference $(ak+b)^{\underline m}$ where $(a,b)=(4,0)$ in your case. 
Such a rule is not given in Concrete Mathematics, however you can find it in Boole's Calculus if Finite Differences, on page 68, eq.(6), it reads
$\sum (ak+b)^{\underline m} = \frac{(ak+b)^{\underline m+1}}{a(m+1)}+C,$
or equivalently
$\Delta (ak+b)^{\underline m}=am(ak+b)^{\underline m-1}.$
