# Evaluate a sum which almost looks telescoping but not quite:$\sum_{k=2}^n \frac{1}{k(k+2)}$ [duplicate]

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?

## marked as duplicate by Guy Fsone, Misha Lavrov, Claude Leibovici, Mostafa Ayaz, Parcly TaxelFeb 13 '18 at 11:00

• I bed a dollar that this is a duplicate. – Guy Fsone Feb 10 '18 at 12:36
• "almost telescoping but not quite". Why isn't it telescoping? The subtracted terms are off-set by two instead of just by one but that won't make any difference. – fleablood Feb 10 '18 at 15:48

\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.

$\require{cancel}$ 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?

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.

Hint :

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

$(1/2)\sum_{k=2}^{n}\dfrac{1}{k}=$

$(1/2)(\dfrac{1}{2} + \dfrac{1}{3} +.......\dfrac{1}{n})$ ;

$(1/2)\sum_{k=2}^{n}\dfrac{1}{k+2}=$

$(1/2)(\dfrac{1}{4}+...\dfrac{1}{n} +\dfrac{1}{n+1} + \dfrac{1}{n+2} ).$