Evaluate a sum which almost looks telescoping but not quite:$\sum_{k=2}^n \frac{1}{k(k+2)}$ Suppose I need to evaluate the following sum:
$$\sum_{k=2}^n \frac{1}{k(k+2)}$$
With partial fraction decomposition, I can get it into the following form:
$$\sum_{k=2}^n \left[\frac{1}{2k}-\frac{1}{2(k+2)}\right]$$
This almost looks telescoping, but not quite... so at this point I am unsure of how to proceed.  How can I evaluate the sum from here?

Hint :
$1/2(\sum_{k=2}^{n}\dfrac{1}{k} -\sum_{k=2}^{n}\dfrac{1}{k+2} ):$.
$(1/2)\sum_{k=2}^{n}\dfrac{1}{k}=$
$(1/2)(\dfrac{1}{2} + \dfrac{1}{3} +.......\dfrac{1}{n})$ ;
$(1/2)\sum_{k=2}^{n}\dfrac{1}{k+2}=$
$(1/2)(\dfrac{1}{4}+...\dfrac{1}{n} +\dfrac{1}{n+1} + \dfrac{1}{n+2} ).$
