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.