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**Pinkk** Ah, I think I get it. So using $\displaystyle S_{3}$ in the first problem (since this when terms begin to cancel each other out) you eventually arrive at:

$\displaystyle S_{3}= 3-\frac{2}{4}-\frac{2}{5}=3 - 2(\frac{1}{3+1}+\frac{1}{3+2})$ and then simply substitute $\displaystyle n=3$ in the denominators and I get what you got.

So for the second problem I posted, is this correct?

$\displaystyle S_{n}=\frac{11}{2} - 3(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3})$ and thus leading to:

$\displaystyle \sum_{k=1}^{\infty}\,\,\frac{9}{k(k+3)} = \frac{11}{2}$