TITLE: Evaluate the sum $\sum_{k=0}^{\infty}\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}$?
QUESTION [8 upvotes]: 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.

REPLY [4 votes]: 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
$$
\sum_{k=0}^\infty\frac1{(4k+1)(4k+2)(4k+3)(4k+4)}=\frac14\log(2)-\frac\pi{24}
$$