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?
 A: \begin{align*}
\dfrac{1}{k(k+2)}&=\dfrac{1}{2}\left(\dfrac{1}{k}-\dfrac{1}{k+2}\right)\\
&=\dfrac{1}{2}\left(\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)+\left(\dfrac{1}{k+1}-\dfrac{1}{k+2}\right)\right),
\end{align*}
splitting the sum and doing telescope twice.
A: $\require{cancel}$
HINT 
Look at the first few terms:
$$\frac14\cancel{-\frac18}+\frac16\cancel{-\frac1{10}}\cancel{+\frac18}-\frac1{12}+\cancel{\frac1{10}}-\cdots$$
Do you notice anything particular?
A: One approach you could use is to manually check which terms cancel out. Notice that:
$$ \sum^{n}_{k=2} \dfrac{1}{k} = \dfrac{1}{2} + \dfrac{1}{3} + \left[\dfrac{1}{5} + ... + \dfrac{1}{n}\right]$$
And,
$$ \sum^{n}_{k=2} \dfrac{1}{k+2} = \left[\dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + ... + \dfrac{1}{n}\right] + \dfrac{1}{n+1} + \dfrac{1}{n+2} $$
The terms in the square brackets get cancelled out on subtraction. 
A: Hint :
$1/2(\sum_{k=2}^{n}\dfrac{1}{k} -\sum_{k=2}^{n}\dfrac{1}{k+2} ):$.
$(1/2)\sum_{k=2}^{n}\dfrac{1}{k}=$
$(1/2)(\dfrac{1}{2} + \dfrac{1}{3} +.......\dfrac{1}{n})$ ;
$(1/2)\sum_{k=2}^{n}\dfrac{1}{k+2}=$
$(1/2)(\dfrac{1}{4}+...\dfrac{1}{n} +\dfrac{1}{n+1} + \dfrac{1}{n+2} ).$
