<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] The partial sum decomposition of the term of your serie is : $\frac{\frac{1}{3}}{n}-\frac{\frac{1}{3}}{n+3}$. You recognise telescoping series : $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}=\sum_{n=1}^{\infty} \frac{\frac{1}{3}}{n}-\frac{\frac{1}{3}}{n+3}=\frac{\frac{1}{3}}{1}+\frac{\frac{1}{3}}{2}+\frac{\frac{1}{3}}{3}=\frac{11}{18}$.
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