TITLE: Computing $\sum\limits_{n=1}^q {1\over (kn-1)(kn)(kn+1)}$ as a telescoping series, when $k\geqslant3$?
QUESTION [5 upvotes]: One knows that $$S_1(q)=\sum_{n=1}^q {1\over (n-1)(n)(n+1)} = \sum_{n=1}^q\frac12\left({\frac{1}{(n-1)n}-\frac{1}{n(n+1)}}\right)$$
and the RHS can be easily telescoped. 
The same approach works for $$S_2(q)=\sum_{n=1}^q {1\over (2n-1)(2n)(2n+1)}$$
However, for $$S_3(q)=\sum_{n=1}^q {1\over (3n-1)(3n)(3n+1)}$$ it is impossible to telescope using the same method than in the two cases above. So:

How should  $$S_k(q)=\sum_{n=1}^q {1\over (kn-1)(kn)(kn+1)}$$ where $k\geqslant3$ is an integer, be telescoped?

REPLY [3 votes]: Partial fractions:
$$\frac1{(3n-1)3n(3n+1)}=\frac a{3n-1}+\frac b{3n}+\frac c{3n+1}\implies$$
$$1=3an(3n+1)+b(3n-1)(3n+1)+3cn(3n-1)$$
Now, attach values to $\;n\;$ to get the values $\;a,b,c\;$ :
$$\begin{align*}&n=0\implies&1=-b\\{}\\
&n=\frac13\implies&1=2a\implies a=\frac12\\{}\\
&n=-\frac13\implies&-1=-2c\implies c=\frac12\end{align*}$$
Thus:
$$\frac1{(3n-1)3n(3n+1)}=\frac12\left(\frac1{3n-1}-\frac2{3n}+\frac1{3n+1}\right)$$