# Formula for the nth partial sum of a telescoping series

Find the $n$th term for the sequence of partial sums for the series $$\sum_{n=1}^{\infty} \frac{5}{n(n+3)} =\sum_{n=1}^{\infty} \left(\frac{5}{3n}-\frac{5}{3(n+3)} \right)$$ and find $\lim\limits_{n\rightarrow \infty} s_n$.

The sequence of partial sums looks like this: $$\{s_n\} = \left\{\frac{5}{4}, \frac{7}{4}, \frac{73}{36}, \frac{139}{63}, \frac{1175}{504},\ldots\right\}$$

What's the best way to go about finding a general expression for the $n$th partial sum?

For a general telescoping series like this one you use the same "trick" as for a "usual" telescoping series: $$\sum_{n=0}^N (u_{n+3}-u_n)=\sum_{n=0}^N u_{n+3}-\sum_{n=0}^N u_n=\sum_{n=3}^{N+3} u_n -\sum_{n=0}^N u_n$$ so: $$\sum_{n=0}^N (u_{n+3}-u_n)=\sum_{n=3}^{N} u_n+u_{N+1}+u_{N+2}+u_{N+3} -\left(u_0+u_1+u_2+\sum_{n=3}^N u_n \right)$$ $$\sum_{n=0}^N (u_{n+3}-u_n)=u_{N+1}+u_{N+2}+u_{N+3}-u_0-u_1-u_2$$
HINT: For $$\sum_{i=1}^n\frac{5}{i(i+3)}$$ we get $$\sum_{i=1}^n\frac{5}{i(i+3)}=-5/3\, \left( n+1 \right) ^{-1}-5/3\, \left( n+2 \right) ^{-1}-5/3\, \left( n+3 \right) ^{-1}+{\frac{55}{18}}$$
HINT: Take four (or more) consecutive terms from the beginning and end and observe reductions. $$\frac53\left( \frac11-\color{cyan}{\frac14}+\frac12-\frac15+\frac13-\frac16+\color{cyan}{\frac14}-\frac17+\ldots \right)$$