Sum $\sum \frac{1}{(4k-3)(4k-2)(4k-1)(4k)}$ I am stuck on this problem for quite a while now, and I don't seem any closer to the solution. So, here it is:
$S = 1/4! + 4!/8! + 8!/12! + 12!/16! + ......$ 
I crossed out the factorials first, and it could be easily represented by the general term, 
$T = \frac{1}{(4n-3)(4n-2)(4n-1)(4n)}$
It looked like it could be expressed as the difference of two expressions which could come useful to find the sum, what we call the 'diagonal cancellation',
$T = \frac{1}{3}(\frac{1}{(4n-1)(4n-2)(4n-3)} - \frac{1}{4n(4n-1)(4n-2)})$
but unfortunately it doesn't work. I even unintentionally split it further into subtractions of two more expressions for both terms and so on, which finally leads to a harmonic series, which is certainly not expressible in "closed-form'.
Please help by suggesting a simple math solution, this is merely a class notes illustration and I am not an advanced maths student. 
 A: \begin{align*}
& \frac{1}{(4n-3)(4n-2)(4n-1)(4n)} \\
&= \frac{1}{3}\left(\frac{1}{(4n-1)(4n-2)(4n-3)} - \frac{1}{4n(4n-1)(4n-2)}\right) \\
& = \frac{1}{3\cdot 2}\left( \frac{1}{(4n-2)(4n-3)} - \frac{2}{(4n-1)(4n-2)} + \frac{1}{4n(4n-1)}\right) \\ 
& = \frac{1}{3\cdot 2\cdot 1}\left( \frac{1}{4n-3} - \frac{3}{4n-2} + \frac{3}{4n-1} - \frac{1}{4n}\right) \\ 
\end{align*}
So the sum is \begin{align*}
&\frac{1}{6}\sum_{n=1}^{\infty}\left( \frac{1}{4n-3} - \frac{3}{4n-2} + \frac{3}{4n-1} - \frac{1}{4n}\right) =\\
& =\frac{1}{6}\sum_{n=1}^{\infty}\left( \int_0^1 x^{4n-4} dx -3 \int_0^1 x^{4n-3} dx+ 3\int_0^1 x^{4n-2} dx - \int_0^1 x^{4n-1} dx\right) = \\
& = \frac{1}{6}\sum_{n=1}^{\infty}\left( \int_0^1 x^{4n-4} - 3 x^{4n-3} +3x^{4n-2}  -  x^{4n-1} dx\right) = \\
& \stackrel{*}{=} \frac{1}{6} \int_0^1 \sum_{n=1}^{\infty} \left(x^{4n-4} - 3 x^{4n-3} +3x^{4n-2}  -  x^{4n-1}\right) dx = \\
& = \frac{1}{6} \int_0^1 \frac{1 - 3x + 3x^2 - x^3}{1-x^4} dx = \\ 
& = \frac{1}{6} \int_0^1 \frac{(1-x)^2}{(1+x)(1+x^2)} dx = \cdots = \frac{1}{24}(6\ln2 - \pi)\\ 
\end{align*}
Here the step $\stackrel{*}{=}$ should be justified; the integrand is of the form $x^{4n-4}(1-x)^3$ and this is nonnegative on $[0,1]$ so Fubini-Tonelli is appliable.
