How to calculate: $\sum_{i=1}^n \frac{n}{(3n+i-1)(3n+i)}$. While solving a question that has to do with Darboux sums I reached the following sum:
$$\sum_{i=1}^n \frac{n}{(3n+i-1)(3n+i)}$$
and I didn't know how to do it. but I know its supposed to be $\frac{1}{12}$.

Telescopic series
$$\sum_{i=1}^n \frac{n}{(3n+i-1)(3n+i)}
\sum_{i=1}^n=\frac{1-i}{3 (i+3 n-1)}+\frac{i}{3 (i+3 n)}=
\left(\frac{0}{3 (3 n)}+\frac{1}{3 (1+3 n)} \right)+
\left(\frac{-1}{3 (1+3 n)}+\frac{2}{3 (2+3 n)}\right)+
\left(\frac{-2}{3 (2+3 n)}+\frac{3}{3 (3+3 n)}\right)+
\left(\frac{-3}{3 (3+3 n)}+\frac{4}{3 (4+3 n)}\right)\dots=
\left( \frac{n}{3(4n)}\right)=\frac{1}{12}
$$
