<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] Using partial fraction decomposition 
$$
    \frac{1}{n(n+3)} = \frac{1}{3} \cdot \frac{1}{n} - \frac{1}{3} \cdot  \frac{1}{n+3}
$$
convince yourself that for any $f$, we have
$$
   \sum_{n=1}^{m} \left(f(n)-f(n+3)\right) = f(1) + f(2) + f(3) - f(m+3) - f(m+2) - f(m+1)
$$
and now take the limit of $m \to \infty$ getting
$$
    \sum_{n=1}^\infty \frac{1}{n(n+3)} = \frac{1}{3} \left(1 + \frac{1}{2} + \frac{1}{3} \right) = \frac{11}{18}
$$
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