Evaluate the sum $\sum_{k=0}^{\infty}\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}$? 
Evaluate the series 
  $$\sum_{k=0}^{\infty}\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}=?$$ 

Can you help me ? This is a past contest problem.
 A: Let
\begin{align}S_n & =\sum_{k=0}^{n}\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}\\
& =\sum_{k=0}^{n}\left(\frac{1}{6}\frac{1}{4k+1}-\frac{1}{2}\frac{1}{4k+2}+\frac{1}{2}\frac{1}{4k+3}-\frac{1}{6}\frac{1}{4k+4}\right).
\end{align}
$$A_n=\sum_{k=0}^{n}\left(\frac{1}{4k+1}-\frac{1}{4k+3}\right),$$
$$B_n=\sum_{k=0}^{n}\left(\frac{1}{4k+1}-\frac{1}{4k+2}+\frac{1}{4k+3}-\frac{1}{4k+4}\right),$$ and
$$C_n=\sum_{k=0}^{n}\left(\frac{1}{4k+2}-\frac{1}{4k+4}\right).$$
It is easy check that 
$$S_n=\frac{1}{3}B_n-\frac{1}{6}A_n-\frac{1}{6}C_n.$$
Therefore,
$$\lim_{n\to\infty}S_n=\lim_{n\to\infty}\frac{2B_n-A_n-C_n}{6}=\frac{2\ln 2-\dfrac{\pi}{4}-\dfrac{1}{2}\ln 2}{6}=\frac{1}{4}\ln 2-\frac{\pi}{24}.$$
A: Start with the Mercator series
$$
\sum_{k=0}^\infty\frac{(-1)^k}{k+1}=\log(2)
$$
and Gregory's series
$$
\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}=\frac\pi4
$$
The Heaviside Method yields
$$
\begin{align}
&\frac6{(4k+1)(4k+2)(4k+3)(4k+4)}\\
&=\frac1{4k+1}-\frac3{4k+2}+\frac3{4k+3}-\frac1{4k+4}\\
&=\color{#C00000}{\left(\frac2{4k+1}-\frac2{4k+2}+\frac2{4k+3}-\frac2{4k+4}\right)}\\
&-\,\color{#00A000}{\left(\frac1{4k+1}-\frac1{4k+3}\right)}-\color{#0000FF}{\left(\frac1{4k+2}-\frac1{4k+4}\right)}
\end{align}
$$
Note that the parts in red, green, and blue are all $O\left(\frac1{k^2}\right)$, so their sums converge absolutely.
Thus,
$$
\begin{align}
\sum_{k=0}^\infty\frac6{(4k+1)(4k+2)(4k+3)(4k+4)}
&=\color{#C00000}{2\log(2)}-\color{#00A000}{\frac\pi4}-\color{#0000FF}{\frac12\log(2)}\\
&=\frac32\log(2)-\frac\pi4
\end{align}
$$
Dividing by $6$ yields
$$
\sum_{k=0}^\infty\frac1{(4k+1)(4k+2)(4k+3)(4k+4)}=\frac14\log(2)-\frac\pi{24}
$$
A: This is Problem 2 of Day 1 of the 2010 International Mathematics Competition.  If you go to the competition website, you can find several solutions, but here is a solution I came up with using the Beta function:

We have the identity ** $$\frac{3!}{(4l+1)(4l+2)(4l+3)(4l+4)}=\int_{0}^{1}x^{4l}(1-x)^{3}\text{d}x.$$  Thus we may write
  $$\sum_{l=0}^{\infty}\frac{1}{(4l+1)(4l+2)(4l+3)(4l+4)}=\sum_{l=0}^{\infty}\frac{1}{3!}\int_{0}^{1}x^{4l}(1-x)^{3}\text{d}x.$$ Rearranging the order of summation and integration yields $$\frac{1}{6}\int_{0}^{1}\left(\sum_{l=0}^{\infty}x^{4l}\right)(1-x)^{3}\text{d}x=\frac{1}{6}\int_{0}^{1}\frac{(1-x)^{3}}{1-x^{4}}\text{d}x=\frac{1}{6}\int_{0}^{1}\frac{(1-x)^{2}}{(1+x)(1+x^{2})}\text{d}x.$$ Using partial fractions to split up the integrand we then have $$\frac{1}{6}\int_{0}^{1}\frac{2}{1+x}-\frac{1+x}{1+x^{2}}\text{d}x=\frac{1}{3}\int_{0}^{1}\frac{1}{1+x}-\frac{1}{6}\int_{0}^{1}\frac{1}{1+x^{2}}\text{d}x-\frac{1}{6}\int_{0}^{1}\frac{x}{1+x^{2}}\text{d}x$$ and evaluating the integrals yields $$=\frac{1}{3}\log2-\frac{1}{6}\frac{\pi}{4}-\frac{1}{12}\log2=\frac{1}{4}\log2-\frac{\pi}{24}$$

** This identity follows from the fact that $\text{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ where $\text{B}(x,y)$ is the Beta Function, but it also follows directly from expanding and integrating since $l$ is an integer.  Essentially, this integral encodes the partial fraction decomposition.  
