Infinite Series using Falling Factorials

[WittyName]

Homework Statement

Determine $\sum_{k=0}^\infty \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}$.

Homework 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 $m\neq-1$.

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.

 

[mighty2000]

Thank you.Your approach is on the right track, but there is a mistake in your calculation. The series you have written is not equivalent to the given series. To see this, try evaluating the first few terms of both series and you will see that they are not the same.

The correct approach is to use partial fractions to rewrite the given series as a sum of simpler series. Specifically, we can write:

$$\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)} = \frac{1}{6}\left(\frac{1}{4k+1}-\frac{1}{4k+2}-\frac{1}{4k+3}+\frac{1}{4k+4}\right)$$

Then we can split the original series into four separate series and evaluate each one individually. Using the result you already found for the series involving (4k)_{-4}, we have:

$$\sum_{k=0}^\infty \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)} = \frac{1}{6}\left(\lim_{n\to\infty}\sum_{k=0}^n (4k)_{-4} - \lim_{n\to\infty}\sum_{k=0}^n (4k+1)_{-4} - \lim_{n\to\infty}\sum_{k=0}^n (4k+2)_{-4} + \lim_{n\to\infty}\sum_{k=0}^n (4k+3)_{-4}\right)$$

Evaluating each of these separately, we get:

$$\sum_{k=0}^\infty \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)} = \frac{1}{6}\left(\frac{1}{18} - \frac{1}{162} - \frac{1}{486} + \frac{1}{486}\right) = \frac{1}{18}$$

which is the correct answer.