Find $S_n$ summation.

Question

What is the formula for $S_n$ where $$S_n = \sum_{i=1}^{n} \frac{1}{(n)(n+1)(n+2)(n+3)}?$$

Can we do partial fractions for solving ?

Answer 1

$$\sum_{k=1}^n\frac{1}{k(k+1)(k+2)(k+3)}=\sum_{k=1}^n\left(\frac{1}{k}-\frac{1}{k+1}\right)\left(\frac{1}{k+2}-\frac{1}{k+3}\right)=$$ $$=\sum_{k=1}^n\left(\frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+2}\right)-\frac{1}{3}\left(\frac{1}{k}-\frac{1}{k+3}\right)-\left(\frac{1}{k+1}-\frac{1}{k+2}\right)+\frac{1}{2}\left(\frac{1}{k+1}-\frac{1}{k+3}\right)\right)=$$ $$=\frac{1}{6}\sum_{k=1}^n\left(\frac{1}{k}-\frac{1}{k+3}\right)-\frac{1}{2}\sum_{k=1}^n\left(\frac{1}{k+1}-\frac{1}{k+2}\right)=$$ $$=\frac{1}{6}\left(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{n+2}\right)=$$ $$=\frac{n(n^2+6n+11)}{18(n+1)(n+2)(n+3)}.$$

Answer 2

\begin{align} {\rm S_n} &= \sum_{i=1}^{n} \frac{1}{(n)(n+1)(n+2)(n+3)}\\ &= \frac 12 \sum_{i=1}^{n} \frac{(n^2+3n+2)-(n^2+3n)}{(n)(n+1)(n+2)(n+3)}\\ &=\frac 12 \sum_{i=1}^{n} \frac{(n+1)(n+2)-n(n+3)}{(n)(n+1)(n+2)(n+3)}\\ &=\frac 12 \sum_{i=1}^{n}\left( \frac{1}{n(n+3)} - \frac{1}{(n+1)(n+2)} \right)\\ &=\frac 12 \sum_{i=1}^{n}\frac{1}{n(n+3)} -\frac 12 \sum_{i=1}^{n} \frac{1}{(n+1)(n+2)} \\ &=\frac 16 \sum_{i=1}^{n}\frac{(n+3)-(n)}{n(n+3)} -\frac 12 \sum_{i=1}^{n} \frac{(n+2)-(n+1)}{(n+1)(n+2)} \\ &=\frac 16 \sum_{i=1}^{n}\left( \frac{1}{n} - \frac 1{n+3} \right) -\frac 12 \sum_{i=1}^{n} \left( \frac{1}{(n+1)} - \frac 1{n+2}\right)\\ \end{align}

...telescoping?

Answer 3

Sure, you can do partial fractions. Write $$\frac 1{n(n+1)(n+2)(n+3)}=\frac an+\frac b{n+1}+\frac c{n+2}+\frac d{n+3}$$ Find $a,b,c,d$ and the sum should telescope nicely.