Calculating $\frac{1}{1\cdot 2\cdot 3}+\frac{1}{5\cdot 6\cdot 7}+\frac{1}{9\cdot 10\cdot 11}+\cdots$ I found this question on an old exam paper - UK GCE A-Level (1972) - equivalent to university entrance level in most countries I believe. The method may be "standard" but has left me stumped. Maybe I am missing something obvious. Can someone give me a hint rather than a full worked solution?

Question
Calculate: $$\dfrac{1}{1\cdot 2\cdot 3}+\dfrac{1}{5\cdot 6\cdot 7}+\dfrac{1}{9\cdot 10\cdot 11}+\cdots$$

What do I notice?
It is an infinite series, so one of Geometric, Maclaurin, Taylor Series might be useful. The sum converges because each term is less than geometric series with ratio (0.5).
The terms are formed from "truncated" factorials (my expression)
So the series can be rewritten
$$\frac{0!}{3!}+\frac{4!}{7!}+\frac{8!}{11!}+\cdots$$
There are three successive positive integers in the denominators of each term in the original series and the multiples of 4 are missing from the denominators.
The integers "within" the factorials in the numerator and denominator are (arithmetically) increasing by 4.
Because it is an infinite series I can't hope to "group" the terms by finding common multiples.
So I get stuck.
Then I cheat and put: $\displaystyle\sum \frac{(4k-4)!}{(4k-1)!}$ into Wolfram Alpha.
The answer $\frac{\ln(2)}{4}$, pops out. So I feel an approach to solution might have something to do with the Maclaurin expansion of $\ln(1+x)$ but I can't get anywhere with this.
Any hints would be gratefully received.
Thanks,
Clive
 A: I have a suspicion that the following method would be more like the one expected of the candidates for this exam.
First we decompose into partial fractions, so, as given already, $$S=\frac 12\sum_{r=0}^{\infty}\left(\frac{1}{4k+1}-\frac{2}{4k+2}+\frac{1}{4k+3}\right)$$
Then we start by writing this out explicitly, so that $$2S=\left(\frac 11-\frac 22+\frac 13\right)+\left(\frac 15-\frac 26+\frac 17\right)+\left(\frac 19-\frac{2}{10}+\frac{1}{11}\right)+...$$
Then we systematically add in and subtract terms, so $$2S=\left(\frac 11-\frac 12+\frac 13-\frac 14\right)+\color{red}{\left(-\frac 12+\frac 14\right)}+\left(\frac 15-\frac 16+\frac 17-\frac 18\right)+\color{red}{\left(-\frac 16+\frac 18\right)}+\left(\frac 19-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}\right)+\color{red}{\left(-\frac{1}{10}+\frac{1}{12}\right)}+...$$
So $$2S=\ln 2-\color {red}{\frac 12\ln 2}$$
Then $$S=\frac 14\ln 2$$
I don't think the integration method as shown by @Dr. MV was expected to be known by those students...
A: We can write the general term of the series as 
$$\frac{1}{(4k+1)(4k+2)(4k+3)}=\frac{1}{2}\left(\frac{1}{4k+1}-\frac{2}{4k+2}+\frac{1}{4k+3}\right)$$
Then, noting that $\int_0^1 x^{4k}\,dx=\frac{1}{4k+1}$, we have
$$\begin{align}
\sum_{n=0}^\infty\frac{1}{2}\left(\frac{1}{4k+1}-\frac{2}{4k+2}+\frac{1}{4k+3}\right)&=\sum_{n=0}^\infty\frac12 \int_0^1 (x^{4k}-2x^{4k+1}+x^{4k+2})\,dx\\\\
&=\frac12 \int_0^1 \frac{(1-2x+x^2)}{1-x^4}\,dx\\\\
&=-\frac12\int_0^1 \frac{x-1}{(x^2+1)(x+1)}\,dx
\end{align}$$
Can you finish now?
