<s>[INST] How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$? How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$ ?
I know the sum converges because it is a positive sum for every $n$ and it is smaller than $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ that converges and equals $1$. I need a direction...
 [/INST] \begin{align*}
  \frac{1}{n(n+3)} &= \frac{1}{3n}-\frac{1}{3(n+3)} \\
  \sum_{n=1}^{\infty} \frac{1}{n(n+3)} &=
  \sum_{n=1}^{\infty}
   \left[ \frac{1}{3n}-\frac{1}{3(n+3)} \right] \\
  &=\lim_{N\to \infty}
    \left[
      \sum_{n=1}^{N} \frac{1}{3n}-\sum_{n=1}^{N} \frac{1}{3(n+3)}
    \right] \\
  &=\sum_{n=1}^{3} \frac{1}{3n}-
    \lim_{N\to \infty} \sum_{n=1}^{3} \frac{1}{3(N+n)} \\
  &=\frac{11}{18}
\end{align*}
</s>