TITLE: Is this sum related to the Gregory's limit?
QUESTION [6 upvotes]: Compute the series
$$\sum_{k=0}^{\infty}\left(\frac{1}{4k+1} - \frac{1}{4k+2}\right)$$

REPLY [7 votes]: And more generally, if $a+b+c+d=0$, 
$$\sum_{k=0}^\infty \left(\dfrac{a}{4k+1} + \dfrac{b}{4k+2}+\dfrac{c}{4k+3}+\dfrac{d}{4k+4}\right) = \dfrac{a-c}{8} \pi + \dfrac{a+c-2d}{4} \ln(2) $$
EDIT: The generalization is that for positive integers $n$,
if $\sum_{j=0}^{n-1} a_j = 0$ then
$$ \sum_{k=1}^\infty \sum_{j=0}^{n-1} \dfrac{a_j}{nk-j} = -\frac{1}{n} \sum_\omega 
\sum_{j=0}^{n-1} a_j \omega^j \ln(1-\omega)$$
where the sum is over all $n$'th roots of unity except $1$.