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 $\frac{1}{12}$.
 A: Hint : Consider $$\frac{n}{3n+i-1}-\frac{n}{3n+i}=\frac{n}{(3n+i-1)(3n+i)}$$
The only terms which do not cancel out are $\frac{n}{3n+i-1}$ for $i=1$ 
and $\frac{n}{3n+i}$ for $i=n$ giving $\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$.
A: Since one of the tag's is "Riemann Sum," I assume that the problem is to find 
$$\lim_{n\to \infty}\sum_{1}^n\frac{n}{(3n+i)(3n+i-1)}$$
by interpreting the sum as a Riemann sum.  To that end, we write
$$\begin{align}
\lim_{n\to \infty}\sum_{1}^n\frac{n}{(3n+i)(3n+i-1)}&=\lim_{n\to \infty}\sum_{1}^n\frac{1}{(3+i/n)(3+(i-1)/n)}\left(\frac1n\right)\\\\
&= \int_0^1\frac{1}{(3+x)^2}\,dx\\\\
&=\left.\left(-\frac{1}{3x+1}\right)\right|_0^1\\\\
&=\frac1{12}
\end{align}$$
as expected!
A: Telescopic series
$$\sum_{i=1}^n \frac{n}{(3n+i-1)(3n+i)}
\sum_{i=1}^n=\frac{1-i}{3 (i+3 n-1)}+\frac{i}{3 (i+3 n)}=
\left(\frac{0}{3 (3 n)}+\frac{1}{3 (1+3 n)} \right)+
\left(\frac{-1}{3 (1+3 n)}+\frac{2}{3 (2+3 n)}\right)+
\left(\frac{-2}{3 (2+3 n)}+\frac{3}{3 (3+3 n)}\right)+
\left(\frac{-3}{3 (3+3 n)}+\frac{4}{3 (4+3 n)}\right)\dots=
\left( \frac{n}{3(4n)}\right)=\frac{1}{12}
$$
A: Notice, the partial fractions $$\frac{n}{(3n+i-1)(3n+i)}=\frac{n}{3n+i-1}-\frac{n}{3n+i}$$
Now, we have 
$$\sum_{i=1}^{n}\frac{n}{(3n+i-1)(3n+i)}$$
$$=\sum_{i=1}^{n}\left(\frac{n}{3n+i-1}-\frac{n}{3n+i}\right)$$
$$=n\sum_{i=1}^{n}\left(\frac{1}{3n+i-1}-\frac{1}{3n+i}\right)$$
$$=n\left(\left(\frac{1}{3n}-\frac{1}{3n+1}\right)+\left(\frac{1}{3n+1}-\frac{1}{3n+2}\right)+\ldots +\left(\frac{1}{4n-1}-\frac{1}{4n}\right)\right)$$
$$=n\left(\frac{1}{3n}-\frac{1}{4n}\right)$$
$$=n\left(\frac{1}{12n}\right)=\color{red}{\frac{1}{12}}$$
